We provide oscillations practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. We focus on oscillations skills mastery so, below you will get all questions that are also asking in the competition exam beside that classroom.
List of oscillations Questions
| Question No | Questions | Class |
|---|---|---|
| 1 | A small body of mass 10 gram is making harmonic oscillations along a straight line with a time period of ( frac{pi}{4} ) and the maximum displacement is 10 cm. The energy of oscillation is в. ( 0.16 times 10^{-2} mathrm{J} ) c. ( 0.6 times 10^{-2} J ) D. ( 0.56 times 10^{-2} J ) | 11 |
| 2 | A particle executing simple harmonic motion with an amplitude ( 5 mathrm{cm} ) and a time period 0.2 s. The velocity and acceleration of the particle when the displacement is ( 5 mathrm{cm} ) is A ( .0 .5 pi m s^{-1}, 0 m s^{-2} ) B. ( 0.5 m s^{-1},-5 pi^{2} m s^{-2} ) C. ( 0 m s^{-1},-5 pi^{2} m s^{-2} ) D. ( 0.5 pi m s^{-1},-0.5 pi^{2} m s^{-2} ) | 11 |
| 3 | A particle executing SHM has a maximum speed of ( 30 mathrm{cm} s^{-1} ) and a maximum acceleration of ( 60 mathrm{cms}^{-2} ). The period of oscillation is: A . ( pi ) в. ( frac{pi}{2} s ) c. ( 2 pi ) D. ( frac{2 pi}{15} ) | 11 |
| 4 | If the time period of a pendulum is ( 25 s ) What is the frequency of the pendulum? A ( cdot 4 times 10^{-2} mathrm{Hz} ) в. ( 4 times 10^{-1} mathrm{Hz} ) c. ( 25 times 10^{-2} mathrm{Hz} ) D. ( 2 times 10^{-2} mathrm{Hz} ) | 11 |
| 5 | The time period of oscillation of a particle, that executes SHM, is ( 1.2 s ). The time, starting from mean position, at which its velocity will be half of its velocity at mean position is? A . ( 0.1 s ) B. ( 0.2 s ) ( c .0 .4 s ) D. ( 0.6 s ) | 11 |
| 6 | The circular motion of a particle with constant speed is A. Periodic but not simple harmonic B. Simple harmonic but not periodic C. Periodic and simple harmonic D. Neither periodic nor simple harmonic | 11 |
| 7 | Any oscillation in which the amplitude of the oscillating quantity decreases with time is termed as A. Damped oscillation B. Free oscillation c. Depletion oscillation D. None of these | 11 |
| 8 | The mass of particle executing S.H.M is gm.lf its periodic time is ( pi ) seconds, the value of force constant is:- A. 4 dynes/cm B. ( 4 mathrm{N} / mathrm{cm} ) ( c cdot 4 N / m ) D. 4 dynes/m | 11 |
| 9 | The ratio of time periods of oscillations of situations shown in figures (i) and (ii) is: A .2: 3 B. ( 3: sqrt{2} ) ( c cdot 4: 3 ) D. 1: 1 | 11 |
| 10 | For a particle executing SHM having amplitude ‘a’ the speed of the particle is one half of its maximum speed when its displacement from the mean position is ( A cdot a / 2 ) в. c. ( a frac{sqrt{3}}{2} ) D. 2a | 11 |
| 11 | The frequency of a particle performing SHM is 12 Hz. Its amplitude is ( 4 mathrm{cm} ). Its initial displacement is ( 2 mathrm{cm} ) towards positive extreme position. Its equation for displacement is A. ( x=0.04 cos (24 pi t+pi / 6) ) В. ( x=0.04 sin (24 pi t) ) c. ( x=0.04 sin (24 pi t+pi / 6) ) D. ( x=0.04 cos (24 pi t) ) | 11 |
| 12 | The displacement of a particle executing SHM at any instant ( t ) is ( x= ) ( 0.01 sin 100(t+0.05) ) then its time period will be A . ( 0.06 mathrm{s} ) B. 0.2 c. ( 0.1 mathrm{s} ) D. 0.02 | 11 |
| 13 | A mass of ( 1 mathrm{kg} ) is suspended from a spring of force constant ( 400 mathrm{N} ) executing SHM total energy of the body is ( 2 mathrm{J} ), then maximum acceleration of the spring will be ( mathbf{A} cdot 4 m / s^{2} ) в. ( 40 m / s^{2} ) c. ( 200 m / s^{2} ) D. ( 400 mathrm{m} / mathrm{s}^{2} ) | 11 |
| 14 | The equilibrium position for a static pendulum changes with A. acceleration of the bob B. velocity of the bob c. displacement of the bob D. none of the above | 11 |
| 15 | State whether true or false. The S.I. unit of frequency is Hertz (Hz). A. True B. False | 11 |
| 16 | A particle performing periodic motion satisfies the equation: ( mathbf{A} cdot f(t)=f(t)+T ) B. ( f(t)=f(t)-T ) c. ( f(t)=f(t+T) ) D. None of the above | 11 |
| 17 | A particle performs SHM on a straight line with time period ( T ) and amplitude A. The average speed of the particle between two successive instants, when potential energy and kinetic energy become same is A ( cdot frac{A}{T} ) в. ( frac{4 sqrt{2} A}{T} ) c. ( frac{2 A}{T} ) D. ( frac{2 sqrt{2} A}{T} ) | 11 |
| 18 | The average energy in one time period in simple harmonic motion is. A ( cdot frac{1}{2} m omega^{2} A^{2} ) B ( cdot frac{1}{4} m omega^{2} A^{2} ) ( mathrm{c} cdot m omega^{2} A^{2} ) D. zero | 11 |
| 19 | A person drives the paddle ball by moving his finger up and down at a certain frequency. In this example he is causing A. a damped vibration B. a forced vibration c. a mechanical vibration D. a transnational vinration | 11 |
| 20 | A particle executing simple harmonic motion has an angular frequency of ( 6.28 s^{-1} ) and an amplitude of ( 10 mathrm{cm}, ) the speed when the displacement is ( 6 mathrm{cm} ) from the mean position is: A. ( 50.2 mathrm{cm} / mathrm{s} ) в. ( 60.2 mathrm{cm} / mathrm{s} ) c. ( 70.2 mathrm{cm} / mathrm{s} ) D. ( 80.2 mathrm{cm} / mathrm{s} ) | 11 |
| 21 | The amplitude of a damped oscillator decreases to 0.9times its original magnitude in 5 s. In another 10 s it will decrease to ( alpha ) times its original magnitude, where ( alpha ) equals A. 0.7 B. 0.81 c. 0.729 D. 0.6 | 11 |
| 22 | The total energy of a simple harmonic oscillator is proportional to A. amplitude B. square of amplitude c. frequency D. velocity | 11 |
| 23 | A particle of mass ( m ) is located in a unidimensional potential field, where the potential energy of the particle depends on the coordinates as ( U(x)= ) ( U_{0}(1-sin b x) ; ) where ( U_{0} ) and ( b ) are constant. find the period of small oscillations that the particle performs about the equilibrium position. A ( cdot frac{2 pi}{b^{2}} sqrt{frac{m}{U_{0}}} ) В. ( frac{pi}{b} sqrt{frac{m}{U_{0}}} ) c. ( frac{pi^{2}}{2 b} sqrt{frac{m}{U_{0}}} ) D. ( frac{2 pi}{b} sqrt{frac{m}{U_{0}}} ) | 11 |
| 24 | In the arrangement shown, the solid cylinder of mass ( mathrm{m} ) is slightly rolled to the left and released. It starts oscillating on the horizontal surface without slipping. Then time period of oscillation is ( ^{mathrm{A}} cdot pi sqrt{frac{3 K}{m}} ) в. ( 2 pi sqrt{frac{3 M}{2 K}} ) c. ( _{2 pi} sqrt{frac{2 K}{3 M}} ) D. ( 2 pi sqrt{frac{3 K}{2 m}} ) | 11 |
| 25 | Which of the following functions correctly represent the travelling wave equation for finite values of ( x ) and ( t ) A ( cdot y=x^{2}-t^{2} ) B ( cdot y=cos x^{2} sin t ) c. ( y=log left(x^{2}-t^{2}right)-log (x-t) ) D・ ( y=e^{2 x} sin t ) | 11 |
| 26 | In damped oscillation mass is 1 kg and spring constant ( =100 N / m, ) damping coefficeint ( =0.5 mathrm{kg} s^{-1} . ) If the mass displaced by ( 10 mathrm{cm} ) from its mean position then what will be the value of its mechanical energy after 4 seconds? A . ( 0.67 J ) в. 0.067 Л c. ( 6.7 J ) D. ( 0.5 J ) | 11 |
| 27 | A particle executes simple harmonic motion with a frequency f. The frequency with which the kinetic energy oscillates is ( A cdot y / 2 ) B. ( c cdot 2 y ) D. none of these | 11 |
| 28 | In which of the following there is some loss of energy in the form of heat A. Forced vibrations B. Free vibration c. Damped vibrations D. All | 11 |
| 29 | The piston in the cylinder head of a locomotive has a stroke of 6 m. If the piston executing simple harmonic motion with an angular frequency of 200 rad ( m i n^{-1}, ) its maximum speed is A ( .5 mathrm{ms}^{-1} ) B . ( 10 mathrm{ms}^{-1} ) c. ( 15 mathrm{ms}^{-1} ) D. ( 20 m s^{-1} ) | 11 |
| 30 | When a particle executing SHM oscillates with a frequency ( nu, ) then the kinetic energy of the particle: A. changes periodically with a frequency of ( nu ) B. changes periodically with a frequency of ( 2 nu ) C . changes periodically with a frequency of ( frac{nu}{2} ) D. remains constant | 11 |
| 31 | Find the torque of a force ( overrightarrow{boldsymbol{F}}=-mathbf{3} hat{mathbf{i}}+ ) ( hat{boldsymbol{j}}+boldsymbol{5} widehat{boldsymbol{k}} ) acting at the point ( overrightarrow{boldsymbol{r}}=boldsymbol{7} hat{boldsymbol{i}}+ ) ( mathbf{3} widehat{boldsymbol{j}}+widehat{boldsymbol{k}} ) ( mathbf{A} cdot 14 hat{i}-38 hat{j}+16 hat{k} ) B . ( 4 hat{i}-4 widehat{j}+6 hat{k} ) c. ( $ $-13-16 ) lhat ( {k}-38 ) lhat ( {j}-16 mid ) hat ( {i} $ $ ) D. none of these | 11 |
| 32 | A particle executing SHM has amplitude of ( 4 mathrm{cm} ., ) and its acceleration at a distance of ( 1 mathrm{cm} ) from the mean position is ( 3 mathrm{cm} s^{-2} . ) Its velocity be when it is at a distance of ( 2 mathrm{cm} ) from its mean position is A. ( 2 c m / s ) в. ( 3 mathrm{cm} / mathrm{s} ) c. ( 4 c m / s ) D. ( 6 mathrm{cm} / mathrm{s} ) | 11 |
| 33 | For a particle executing simple harmone motion, the kinetic energy k is given by ( k=k_{0} cos ^{2} omega t . ) The maximum value of potential energy is? ( A cdot k_{0} ) B. zero ( c cdot k_{0} / 2 ) D. not obtainable | 11 |
| 34 | A body is lying on a piston which is executing vertical SHM. Its time period is 2 s. For what value of amplitude, the body will leave the piston: ( A cdot 1 m ) B. 0.248 m c. ( 0.428 mathrm{m} ) D. ( 0.842 mathrm{m} ) | 11 |
| 35 | Acceleration-time graph of a particle executing SHM is as shown in the figure. Select the correct alternative(s) This question has multiple correct options A. Displacement of particle at 1 is negative B. Velocity of particle at 2 is positive C. Potential energy of particle at 3 is maximum D. Speed of particle at 4 is decreasing | 11 |
| 36 | In a non harmonic motion; A. Displacement and amplitude are independent of each other B. Displacement and amplitude are dependent on each other C . restoring force is proportional to displacement D. restoring force is zero | 11 |
| 37 | Consider a spring that exerts the following restoring force: ( boldsymbol{F}=-boldsymbol{k} boldsymbol{x} ) for ( boldsymbol{x}>0 ) and ( boldsymbol{F}=-4 boldsymbol{k} boldsymbol{x} ) for ( boldsymbol{x}<mathbf{0} ) A mass m on a frictionless surface is attached to the spring, displaced to ( x=A ) by stretching the spring and then releasing: This question has multiple correct options A cdot The period of motion will be ( T=frac{3}{2} pi sqrt{frac{m}{k}} ) B. The most negative value, the mass ( m ) can reach will be ( x=-frac{A}{2} ) c. The time taken to move from ( x=A ) to ( x=-frac{A}{sqrt{2}} ) straight away will be equal to ( frac{5 pi}{8} sqrt{frac{m}{k}} ) D. The total energy of oscillations will be ( frac{5}{2} k A^{2} ) | 11 |
| 38 | A particle in S.H.M. has a period of 2 seconds and amplitude of ( 10 mathrm{cm} ) Calculate the acceleration when it is at ( 4 c m ) from its positive extreme position. | 11 |
| 39 | A particle executes simple harmonic motion with an amplitude of ( 4 mathrm{cm} . ) At the mean position the velocity of the particle is ( 10 mathrm{cm} / mathrm{s} ). The distance of the particle from the mean position when its speed becomes ( 5 mathrm{cm} / mathrm{s} ) is A ( cdot sqrt{3} mathrm{cm} ) B. ( sqrt{5} mathrm{cm} ) c. ( pm 2(sqrt{3}) ) cm D. ( 2(sqrt{5}) mathrm{cm} ) | 11 |
| 40 | Statement 1: In simple harmonic motion, the velocity is maximum when the acceleration is minimum. Statement 2: Displacement and velocity of ( S . H . M ) differ in phase by ( frac{pi}{2} ) | 11 |
| 41 | A plank with a body of mass ( m ) placed on it starts moving straight up according to the law ( y=a(1-cos omega t) ) where ( y ) is the displacement from the initial position, ( omega=11 ) rad/s. What is the force that the body exerts on the plank? | 11 |
| 42 | Vertical displacement of a plank with a body of mass ‘ ( m ) ‘ on it is varying according to law ( boldsymbol{y}=sin (boldsymbol{omega} boldsymbol{t})+ ) ( sqrt{3} cos (omega t) . ) The minimum value of ( omega ) for which the mass just breaks off the plank and the moment this occurs first ( operatorname{after} t=0 ) are given by ( (y ) is positive vertically upwards): A ( cdot sqrt{frac{g}{2}}, sqrt{frac{2}{g}} frac{pi}{6} ) В ( cdot frac{g}{sqrt{2}} frac{2}{3}, sqrt{frac{pi}{g}} ) c. ( sqrt{frac{g}{2}} frac{pi}{3}, sqrt{frac{2}{g}} ) D. ( sqrt{2 g}, sqrt{frac{2 pi}{3 g}} ) | 11 |
| 43 | A particle performing SHM takes time equal to T (time period of SHM) in consecutive appearances at a particular point. This point is A. An extreme position B. The mean position C. Between positive extreme and mean position D. Between negative extreme and mean position | 11 |
| 44 | A body of mass ( 0.1 k g ) is executing simple harmonic motion according to the equation ( boldsymbol{x}=mathbf{0 . 5} cos left(mathbf{1 0 0} boldsymbol{t}+frac{mathbf{3} boldsymbol{pi}}{mathbf{4}}right) ) metre. Find: (i) the frequency of oscillation, (ii) initial phase, (iii) maximum velocity, (iv) maximum acceleration, (v) total energy. | 11 |
| 45 | A particle executes S.H.M with a period of 6 seconds. If the maximum speed is ( 3.14 mathrm{cm} / mathrm{sec}, ) then what is its amplitude? A. ( 3 mathrm{cm} ) B. ( 5 mathrm{cm} ) ( c cdot 6 mathrm{cm} ) D. ( 9 mathrm{cm} ) | 11 |
| 46 | A simple harmonic motion along the ( x- ) axis has the following properties: amplitude ( =0.5 m, ) the time to go from one extreme position to other is, 2 s and ( boldsymbol{x}=mathbf{0 . 3} boldsymbol{m} ) at ( boldsymbol{t}=mathbf{0 . 5} boldsymbol{s} . ) The general equation of the simple harmonic motion is A ( cdot x=(0.5 m) sin left[frac{pi t}{2}+8^{circ}right] ) B. ( x=(0.5 m) sin left[frac{pi t}{2}-8^{circ}right] ) ( mathbf{c} cdot_{x}=(0.5 m) cos left[frac{pi t}{2}+8^{circ}right] ) D. ( x=(0.5 m) cos left[frac{pi t}{2}-8^{circ}right] ) | 11 |
| 47 | A small block oscillates back and forth on a smooth concave surface of radius ( boldsymbol{R} ). The time period of small oscillation of the block is : ( ^{mathbf{A}} cdot T=2 pi sqrt{frac{R}{g}} ) В ( cdot T=2 pi sqrt{frac{2 R}{g}} ) c. ( T=2 pi sqrt{frac{R}{2 g}} ) D. none of these | 11 |
| 48 | A particle of mass ( m ) is allowed to oscillate on a smooth parabola: ( boldsymbol{x}^{2}= ) ( 4 a y, a>1 ) as shown in the figure. The angular frequency ( (omega) ) of small oscillations is A ( cdot omega=sqrt{frac{g}{4 a}} ) B. ( omega=sqrt{frac{g}{2 a}} ) c. ( omega=sqrt{frac{2 g}{a}} ) ( omega=sqrt{frac{g}{a}} ) | 11 |
| 49 | The variation in potential energy of a harmonic oscillator is as shown in the figure. The spring constant is A ( cdot 1 times 10^{2} mathrm{Nm}^{-1} ) B. ( 1.5 times 10^{2} N m^{-1} ) c. ( 0.0667 times 10^{2} N m^{-1} ) D. ( 3 times 10^{2} mathrm{Nm}^{-} ) | 11 |
| 50 | The tension of a stretched string is increases by ( 44 % ). In order to keep its frequency of vibration same its length must be increased by. A . ( 10 % ) B. 20% c. ( 15 % ) D. ( 25 % ) | 11 |
| 51 | Figure shows a small magnetised needle ( P ) placed at a point ( O . ) The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle ( Q ) In which configuration the system is not in equilibrium? | 11 |
| 52 | A small block oscillates back and forth on a smooth concave surface of radius R. Find the time period of small oscillations. A ( cdot pi sqrt{frac{2 R}{g}} ) в. ( 2 pi sqrt{frac{R}{3 g}} ) c. ( 2 pi sqrt{frac{2 R}{3 g}} ) D. ( 2 pi sqrt{frac{R}{g}} ) | 11 |
| 53 | Which of the following characteristics must remain constant for undamped oscillations of the particle? A. acceleration B. phase c. amplitude D. velocity | 11 |
| 54 | A wedge of mass ( m ), having a smooth semi circular part of radius R is resting on smooth horizontal surface.Now a particle of mass ( m / 2 ) is released from the top point of the semicircular part. Then This question has multiple correct options A. The maximum displacement of the wedge will be R B. The maximum displacement of the wedge will be ( frac{2 R}{3} ) c. The wedge will perform simple harmonic motion of amplitude D. The wedge will perform oscillatory motion of amplitude ( frac{R}{3} ) | 11 |
| 55 | The masses in figure slide on a frictionless table. ( m_{1} ) but not ( m_{2}, ) is fastened to the spring. If now ( m_{1} ) and ( m_{2} ) are pushed to the left, so that the spring is compressed a distance ( mathrm{d} ) what will be the amplitude of the oscillation of ( m_{1} ) after the spring system is released? ( ^{mathrm{A}} cdot_{A}=(sqrt{frac{m_{1}}{m_{1}+m_{2}}}) ) B. ( A=(sqrt{frac{m_{2}}{m_{1}+m_{2}}}) d ) ( ^{mathbf{c}} cdot_{A}=(sqrt{frac{m_{1}}{m_{1}+m_{2}}}) d ) D. None | 11 |
| 56 | Function ( boldsymbol{x}=boldsymbol{A} sin ^{2} boldsymbol{omega} boldsymbol{t}+boldsymbol{B} cos ^{2} boldsymbol{omega} boldsymbol{t}+ ) ( C sin omega t cos omega t ) represents SHM A. For any value of ( A, B ) and ( C ) (except ( C=0 ) ) B. ( A=-B, C=2 B, ) amplitude ( =mid B sqrt{2} ) c. ( A=B ; C=0 ) D. ( A=B ; C=2 B, ) amplitude ( =|B| ) | 11 |
| 57 | A particle moves along the ( x ) -axis according to: ( boldsymbol{x}=boldsymbol{A}[mathbf{1}+boldsymbol{s} boldsymbol{i} boldsymbol{n} boldsymbol{omega} boldsymbol{t}] . ) What distance does it travel between ( t=0 ) and ( boldsymbol{t}=mathbf{2 . 5} boldsymbol{pi} / boldsymbol{omega} ? ) A. ( 4 A ) B. 6A ( c cdot 5 A ) D. None | 11 |
| 58 | A thin fixed ring of radius ( 1 mathrm{m} ) has a positive charge ( 1 times 10^{-5} mathrm{C} ) uniformly distributed over it. A particle of mass ( 0.9 mathrm{g} ) and having a negative charge of ( 1 times 10^{-6} mathrm{C} ) is placed on the axis at a distance of ( 1 mathrm{cm} ) from the centre of the ring. Calculate the time period of oscillations. A . 0.5 secs B. 9.28 secs c. 0.628 secs D. 0.1 secs | 11 |
| 59 | A force ( F=-4 x-8 ) is acting on a block where ( x ) is position of block in meter The energy of oscillation is 32 J, the block oscillate between two points Position of extreme position is: ( A cdot 6 ) B. 0 ( c cdot 4 ) D. 3 | 11 |
| 60 | If two SHMs of different amplitudes are added together, the resultant SHM will be a maximum if the phase difference between them is ( mathbf{A} cdot pi / 2 ) в. ( pi / 4 ) ( c ) D. ( 2 pi ) | 11 |
| 61 | A body is executing S.H.M. When its displacement from the mean position is ( 4 mathrm{cm} ) and ( 5 mathrm{cm}, ) the corresponding velocity of the body is ( 10 mathrm{cm} / mathrm{sec} ) and 8 ( mathrm{cm} / mathrm{sec} . ) Then the time period of the body is: A ( .2 pi sec ) B . ( pi / 2 ) sec ( c . pi mathrm{sec} ) D. 3pi/2 sec | 11 |
| 62 | In an SHM, restoring force is ( boldsymbol{F}=-boldsymbol{k} boldsymbol{x} ) where ( k ) is force constant, ( x ) is displacement and ( A ) is the amplitude of motion, then the total energy depends upon A. ( k, A ) and ( M ) в. ( k, x, M ) ( c cdot k, A ) D. ( k, x ) | 11 |
| 63 | A highly rigid cubical block A of smal mass M and side L is fixed rigidly onto another cubical block B of same dimensions and of low modulus of rigidly ( eta ) such that lower face of ( A ) completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force is applied perpendicular to one the side face of A. After the force is withdrawn, block A executes small oscillations, the time period of which is given by в. ( 2 pi sqrt{frac{M}{eta L}} ) c. ( 2 pi sqrt{frac{M L}{eta}} ) D. ( sqrt{frac{M eta}{L}} ) | 11 |
| 64 | A particle is executing SHM with amplitude ( A ) and has maximum velocity ( V_{0} . ) Find its speed when it is located at distance of A/2 from mean position. | 11 |
| 65 | Which of the following quantities are always negative? A ( cdot vec{F} cdot vec{r} ) в. ( vec{v} cdot vec{r} ) c. ( vec{a} cdot vec{r} ) D. Both ( A ) and ( C ) | 11 |
| 66 | At ( t=0, ) the displacement of a particle in S.H.M. is half its amplitude. Its initial phase is: A ( cdot frac{pi}{6} ) rad B ( cdot frac{pi}{3} ) rad c. ( frac{2 pi}{3} ) rad D. ( frac{pi}{2} ) rad | 11 |
| 67 | If density (D) acceleration (a) and force (F) are taken as basic quantities,then Time period has dimensions A ( cdot frac{1}{6} ) in ( F ) B. ( -frac{1}{6} ) in ( F ) ( c cdot-frac{2}{3} ) in ( F ) D. All the above are true | 11 |
| 68 | Phase different between the instantaneous velocity and acceleration of particle executing SHM is A . zero в. ( frac{pi}{2} ) ( c . pi ) D. ( 2 pi ) | 11 |
| 69 | A horizontal platform with an object placed on it is executing simple harmonic motion in the vertical direction. The amplitude of oscillation is ( 3.92 times 10^{-3} ) m. What should be the least period of these oscillations, so that the object is not detached from the platform? A. 0.145 sec B. 0.1556 sec c. ( 0.1256 mathrm{sec} ) D. ( 0.1356 mathrm{sec} ) | 11 |
| 70 | The amplitude and the time period in a S.H.M. is ( 0.5 mathrm{cm} ) and 0.4 see respectively If the initial phase is ( r / 2 ) radiam,then the equaction of S.H.H. will be A ( cdot y=0.5 sin 5 / pi ) B. ( y=0.5 sin 4 / pi ) c. ( y=0.5 sin 2.5 / pi ) D. ( y=0.5 cos 5 pi t ) | 11 |
| 71 | Which of the following functions of time represent ( (A) ) simple harmonic, ( (B) ) periodic but not simple harmonic, and ( (C) ) non-periodic motion? Give period for each case of periodic motion ( (omega ) is any positive constant): (a) ( sin omega t-cos omega t ) ( (b) sin ^{3} omega t ) (c) ( 3 cos (pi / 4-2 omega t) ) ( (d) cos omega t+cos 3 omega t+cos 5 omega t ) ( (e) exp left(-omega^{2} t^{2}right) ) (f) ( 1+omega t+omega^{2} t^{2} ) | 11 |
| 72 | A plot of displacement of a particle with time is as shown in the figure. The equation of SHM is A ( . x= ) Asinwt B. ( x= ) Acoswt c. ( x=- ) Acoswt D. ( x=- ) Asinwt | 11 |
| 73 | A point mass oscillates along the ( x ) axis according to the law ( boldsymbol{x}= ) ( x_{0} cos (omega t-pi / 4) . ) If the acceleration of the particle is written as ( a= ) ( A cos (omega t+delta) ) then: A ( . A=x_{0}, delta=-pi / 4 ) B . ( A=x_{0} omega^{2}, delta=-pi / 4 ) C ( . A=x_{0} omega^{2}, delta=-pi / 8 ) D. ( A=x_{0} omega^{2}, delta=3 pi / 4 ) | 11 |
| 74 | Average kinetic energy in one time period of a simple harmonic oscillator whose amplitude is ( A, ) angular velocity ( omega ) and mass ( m, ) is A ( cdot frac{1}{4} m omega^{2} A^{2} ) B ( cdot frac{1}{2} m omega^{2} A^{2} ) ( mathrm{c} cdot m omega^{2} A^{2} ) D. Zero | 11 |
| 75 | A particle is moving with a constant acceleration. Its velocity is reduced to zero in 5 s and it covered a distance of ( 100 mathrm{m} ) in this direction. The distance covered by the particle in the next 5 s is A. Zero B. 250 m c. ( 100 mathrm{m} ) D. 500 ( mathrm{m} ) | 11 |
| 76 | The equation ( x=a sin 2 t+b cos 2 t ) will represent an SHM A. True B. False | 11 |
| 77 | A simple harmonic oscillator has an acceleration of ( 1.25 mathrm{m} / mathrm{s}^{2} ) at ( 5 mathrm{cm} ) from the equilibrium. Its period of oscillation is: A ( frac{4 pi}{5} ) s в. ( frac{5 pi}{2} ) s c. ( frac{2 pi}{5} ) s D. ( frac{2 pi}{25} ) s | 11 |
| 78 | Q Type your question- length ( 4.9 mathrm{m} ). The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by ( 0.2 mathrm{m} ) and released from rest at ( t=0 . ) It then executes simple harmonic motion with angular frequency ( omega=frac{pi}{3} r a d / s ) Simultaneously at ( t=0, ) a small pebble is projected with speed v from point ( mathrm{P} ) at an angle of 45 as shown in the figure. Point ( P ) is at a horizontal distance of 10 ( mathrm{m} ) from ( 0 . ) If the pebble hits the block at ( t=1 mathrm{s}, ) the value of ( mathrm{v} ) is ( left(operatorname{take} g=10 m / s^{2}right) ) ( mathbf{A} cdot sqrt{50} m / s ) B. ( sqrt{51} mathrm{m} / mathrm{s} ) c. ( sqrt{52} mathrm{m} / mathrm{s} ) D. ( sqrt{53} mathrm{m} / mathrm{s} ) | 11 |
| 79 | A particle executes SHM given by the equation ( boldsymbol{x}=mathbf{4} sin (mathbf{2} boldsymbol{pi} boldsymbol{t}+boldsymbol{pi} / mathbf{4}), ) what will be the velocity of the particle at ( t= ) (1/8)th sec; A. velocity = 0 B. velocity = maximum c. velocity = minimum D. The particle’s speed cannot be determined | 11 |
| 80 | A simple harmonic oscillator is of mass ( 0.100 mathrm{kg} . ) It is oscillating with a frequency of ( frac{5}{pi} mathrm{Hz} ). If its amplitude of vibration is ( 5 mathrm{cm}, ) the force acting on the particle at its extreme position is A. 2 N B. 1.5 N ( c cdot 1 mathrm{N} ) D. 0.5 N | 11 |
| 81 | The maximum kinetic energy of a particle of mass 100 g and time period lpi secs, executing SHM at its mean position is given as 50 J. The amplitude of oscillation is A ( . A=50 sqrt{10} mathrm{m} ) В. ( A=5 sqrt{10} m ) c. ( A=50 m ) D. ( A=25 m ) | 11 |
| 82 | Assertion: In forced oscillations, the steady state motion of the particle is simple harmonic. Reason : Then the frequency of particle after the free oscillations die out, is the natural frequency of the particle. A. If both assertion and reason are true and reason is the correct explanation of assertion. B. If both assertion and reason are true and reason is not the correct explanation of assertion. c. If assertion is true but reason is false D. If both assertion and reason are false | 11 |
| 83 | The period of a particle in linear SHM is 8 s. ( A t t=0 ) it is at the mean position. Find the ratio of distance traveled by it in 1 st second and 2 nd second A . 3.2: B. 2.4: c. 1.6: D. 4.2: | 11 |
| 84 | The velocity of a sound wave in ( v ) and the wave energy density is ( E, ) then the amount of energy transferred per unit area per second by the wave in a direction normal to the wave propagation is A. ( E ) в. ( E v ) ( mathrm{c} cdot E^{2} v^{2} ) D. ( sqrt{E v} ) | 11 |
| 85 | The potential energy of a particle executing SHM varies sinusoidally with frequency f. The frequency of oscillation of the particle will be: ( A cdot frac{f}{2} ) в. ( frac{f}{sqrt{2}} ) ( c cdot f ) D. ( 2 f ) | 11 |
| 86 | Potential Energy(U) of a body of unit mass moving in a one-dimension conservative force field is given by ( U= ) ( left(x^{2}-4 x+3right) . ) All units are in S.I. (i) Find the equilibrium positions of the body (ii) Show that oscillations of the body about this equilibrium positions is simple harmonic motion & find its time period. (iii) Find the amplitude of oscillations if speed of the body at equilibrium position is ( 2 sqrt{6} mathrm{m} / mathrm{s} ) | 11 |
| 87 | Assertion There is no difference in the graphs of momentum an velocity of a body performing SHM. Reason Momentum is proportional to velocity of a body. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect and Reason are incorrect | 11 |
| 88 | An object swinging on the end of a string forms a simple pendulum. Some students (and some texts) often cite the simple pendulum’s motion as an example of SHM. That is not quite accurate because the motion is really A. approximately SHM only for small amplitudes B. exactly SHM only for amplitudes that are smaller than a certain value c. approximately SHM for all amplitudes. D. None of the above | 11 |
| 89 | A particle moves in ( x ) -y plane according to the equation ( vec{r}=(hat{i}+hat{j})(A sin omega t+B ) ( cos omega t) . ) Motion of the particle is: A. periodic в. sнм c. along a straight line D. along an ellipse | 11 |
| 90 | A particle is executing simple harmonic motion (SHM) of amplitude ( A ), along the ( x ) -axis, about ( x=0 . ) When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be : A ( cdot frac{A}{2} ) в. ( frac{A}{2 sqrt{2}} ) c. ( frac{A}{sqrt{2}} ) D. | 11 |
| 91 | A block is kept on a horizontal table.The table is undergoing simple harmonic motion of frequency ( 3 H z ) in a horizontal plane.The coefficient of static friction between block and the table surface is ( 0.72 . ) Find the maximum amplitude (in ( mathrm{cm} ) ) of the table at which the block does not slip on the surface. ( Take ( left.g=10 m / s^{2} text { and } pi^{2}=10right) ) | 11 |
| 92 | Two bodies ( mathrm{M} ) and ( mathrm{N} ) of equal mass are suspended from two separate massless spring of force constant ( k_{1} ) and ( k_{2} ) respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of ( mathrm{M} ) to that of ( mathrm{N} ) is? ( mathbf{A} cdot frac{k_{1}}{k_{2}} ) В. ( sqrt{frac{k_{1}}{k_{2}}} ) ( mathbf{c} cdot frac{k_{2}}{k_{1}} ) D. ( sqrt{frac{k_{2}}{k_{1}}} ) | 11 |
| 93 | The displacement of an object attached to a spring and executing simple harmonic motion is given by ( boldsymbol{x}=mathbf{2} times ) ( 10^{-2} ) cos ( pi t ) meters. The time at which the maximum speed first occurs is : A . ( 0.5 mathrm{s} ) B. 0.75 s c. 0.125 s D. ( 0.25 mathrm{s} ) | 11 |
| 94 | A particle of mass m free to move in the x-y plane is subjected to a force whose components are ( boldsymbol{F}_{boldsymbol{x}}=-boldsymbol{k} boldsymbol{y} ) and ( boldsymbol{F}_{boldsymbol{y}}= ) ( -k y, ) where ( k ) is a constant. The particle is released when ( t=0 ) at the point (2,3) Prove that the subsequent motion is simple harmonic along the straight line ( 2 y-3 x=0 ) | 11 |
| 95 | Which of the following statements is/are correct (i) The distance between two consecutive compressions is called frequency of sound wave (ii) Sound propagates as pressure variations (iii) Compressions are narrower in a high pitched sound ( A cdot( ) i) only B . (ii) only c. (iii) only D. (ii) and (iii) | 11 |
| 96 | The tendency of one object to force another adjoining or interconnected object into vibration motion is referred to as a A. forced vibration B. damped vibration c. loudness D. pitch | 11 |
| 97 | The differential equation representing the SHM of a particule is ( frac{mathbf{9} d^{2} boldsymbol{y}}{d t^{2}}+mathbf{4} boldsymbol{y}= ) 0. The time period of the particle is given by : A ( cdot frac{pi}{3} ) sec B . ( pi ) sec c. ( frac{2 pi}{3} ) sec D. ( 3 pi ) sec | 11 |
| 98 | A particle is executing S.H.M with a frequency ‘f’, the frequency with which is K.E. oscillates is? ( mathbf{A} cdot mathbf{f} ) B. ( f / 2 ) c. ( 2 f ) D. ( 4 f ) | 11 |
| 99 | A highly rigid cubical block A of small mass ( mathrm{M} ) and side ( mathrm{L} ) is fixed rigidly on to another cubical block ( mathrm{B} ) of the same dimensions and of low modulus of rigidity ( eta ) such that the lower face of ( A ) completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of a sides faces of A. After the force is withdrawn, block A executes small oscillations, the time period of which is given by ( mathbf{A} cdot 2 pi sqrt{M eta L} ) в. ( 2 pi sqrt{M eta / L} ) c. ( 2 pi sqrt{M L / eta} ) D. ( 2 pi sqrt{M / eta L} ) | 11 |
| 100 | In SHM, select the wrong statement, where ( F ) is the force, ( a ) is the acceleration and ( v ) is the velocity of the particle in SHM. A ( cdot vec{F} times vec{v} ) is a null vector В . ( |vec{F} times vec{a}|=0 ) (always) c. ( vec{F} times vec{a}<0 ) D. ( vec{a} cdot vec{x}<0 ) | 11 |
| 101 | The periodic time of a body executing SHM is ( 4 s ). After how much interval from time ( t=0, ) its displacement will be half of its amplitude? A . ( 1 / 2 s ) B. ( 1 / 3 s ) c. ( 1 / 4 s ) D. ( 1 / 6 s ) | 11 |
| 102 | The coefficient of friction between two blocks of masses ( 1 mathrm{kg} ) and ( 4 mathrm{kg} ) as shown in figure is ( mu ) and the horizontal plane is smooth. If the system is slightly displaced and released it will execute S.H.M. The maximum amplitude if the upper block does not slip relative to lower block will be (k is spring constant) ( A cdot frac{5 mu g}{k} ) в. ( frac{mu g}{k} ) ( c cdot frac{3 mu g}{k} ) D. ( frac{2 mu g}{k} ) | 11 |
| 103 | For the block under SHM shown in the figure, which of the following quantities is not constant? vertica oscillation A. Amplitude B. Frequency c. Period D. Position of block E. Total mechanical energy of the block | 11 |
| 104 | The differential equation for a particle of mass ( 2 mathrm{kg} ) executing ( mathrm{SHM} ) is ( 2 frac{d^{2} x}{d t^{2}}+ ) ( x=0 . ) If amplitude of oscillation is 3 ( mathrm{cm}, ) then maximum velocity of the particle will be A. ( 3 sqrt{0.5} mathrm{cm} / mathrm{s} ) B. ( sqrt{0.5} mathrm{cm} / mathrm{s} ) c. ( sqrt{0.5} / 3 mathrm{cm} / mathrm{s} ) D. ( 2 sqrt{0.5} mathrm{cm} / mathrm{s} ) | 11 |
| 105 | The equation of motion of particle is given by dp ( / mathrm{dt}+mathrm{m} omega^{2} mathrm{y}=0 ) where p is momentum and y is its position. Then the particle : A. moves along a straight line B. moves along a parabola c. executes simple harmonic motion D. falls freely under gravity | 11 |
| 106 | A paricle executes SHM with time period of 4 s. find the minimum time interval in which the velocity of the particle changes by an amount equal to its maximum velocity: A ( cdot frac{1}{2} s ) B. ( frac{2}{3} ) s c. ( frac{9}{3} ) s D. ( frac{5}{3} ) s | 11 |
| 107 | Assertion (A): In damped vibrations, amplitude of oscillation decreases Reason (R): Damped vibrations indicate loss of energy due to air resistance A. Both A and R are true and R is the correct explanation of A B. Both A and R are true and R is not the correct explanation of C. A is true and ( R ) is false D. A is false and R is true | 11 |
| 108 | A system is oscillating with undamped simple harmonic motion.Then the This question has multiple correct options A. average total energy per cycle of the motion is its maximum kinetic energy B. average total energy per cycle of the motion is ( frac{1}{sqrt{2}} ) times its maximum kinetic energy C root mean square velocity is ( frac{1}{sqrt{2}} ) times its maximum velocity D. mean velocity is ( frac{1}{2} ) of maximum velocity | 11 |
| 109 | thingy hanging from a spring. The system was set vibrating by pulling the thingy down below its equilibrium position and then letting it go from rest. If the initial displacement is doubled what happens to the maximum kinetic energy of the thing? A. It is unchanged. B. It is doubled c. It is increased by a factor of ( 4 . ) D. We can’t tell from the information provided | 11 |
| 110 | A simple harmonic motion has an amplitude ( A ) and time period ( T . ) Find the time required by it to travel directly from ( boldsymbol{x}=-frac{boldsymbol{A}}{sqrt{mathbf{2}}}, ) to ( boldsymbol{x}=frac{boldsymbol{A}}{sqrt{mathbf{2}}} ) | 11 |
| 111 | All oscillatory motions are A. periodic B. linear c. rotatory D. curvilinear | 11 |
| 112 | Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by ( 45^{circ}, ) then : This question has multiple correct options A. The resultant amplitude is ( (1+sqrt{2}) a ) B. The phase of the resultant motion relative to the first is ( 90^{circ} ) c. The energy associated with the resulting motion is ( (3+2 sqrt{2}) ) times the energy associated with any single motion D. The resulting motion is not simple harmonic | 11 |
| 113 | A particle of mass ( M ) is executing oscillations about the origin on the ( x ) axis. Its potential energy is ( |boldsymbol{U}|=boldsymbol{K}left|boldsymbol{x}^{2}right| ) where ( K ) is a positive constant. If the amplitude of oscillations is ( a ), then its period ( boldsymbol{T} ) is A. proportional to ( 1 / sqrt{a} ) B. independent of ( a ) c. proportional to ( sqrt{a} ) D. proportional to ( a^{1 / 2} / 2 ) poorional | 11 |
| 114 | The maximum acceleration of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible when A. amplitude of oscillation is doubled while frequency remains constant B. amplitude is doubled while frequency is halved c. frequency is doubled while amplitude is halved D. frequency is doubled while amplitude remains constant | 11 |
| 115 | In periodic motion, the displacement is A. directly proportional to the restoring force B. inversely proportional to the restoring force c. independent of restoring force D. independent of any force acting on the particle | 11 |
| 116 | A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force ( boldsymbol{F} ) sinw. If the amplitude of the particle is maximum for ( omega=omega_{1} ) and the energy of the particle is maximum for ( omega=omega_{2} ) then (where ( omega_{0} ) natural frequency of oscillation of particle) ( mathbf{A} cdot omega_{1}=omega_{0} ) and ( omega_{2} neq omega_{0} ) B . ( omega_{1}=omega_{0} ) and ( omega_{2}=omega_{0} ) C ( cdot omega_{1} neq omega_{0} ) and ( omega_{2}=omega_{0} ) D. ( omega_{1} neq omega_{0} ) and ( omega_{2} neq omega_{0} ) | 11 |
| 117 | A body of mass ( mathrm{M} ) is situated in a potential field ( boldsymbol{U}(boldsymbol{x})=boldsymbol{U}_{0}(mathbf{1}-cos boldsymbol{d} boldsymbol{x}) ) where ( U_{0} ) and ( d ) are constants. The time period of small oscillations will be в. ( 2 pi sqrt{frac{M}{U_{0} d^{2}}} ) ( ^{mathbf{C}} cdot 2 pi sqrt{frac{U_{0} d^{2}}{M}} ) D. ( 2 pi sqrt{frac{U_{0}}{M d^{2}}} ) | 11 |
| 118 | ILLUSTRATION 33.9 To find the value of ‘g’ using simple pendulum. T = 2.00 s;l = 1.00 m was measured. Estimate maximum permissible error in ‘g’. Also find the value of ‘g’. (use it = 10) 4726 | 11 |
| 119 | The number of independent constituent simple harmonic motions yielding a resultant displacement equation of the periodic motion as ( boldsymbol{y}= ) ( 8 sin ^{2}left(frac{t}{2}right) sin (10 t) ) is: ( A cdot 8 ) B. 6 ( c cdot 4 ) ( D .3 ) | 11 |
| 120 | A particle executes SHM given by the equation ( x=5 sin (3.14 t), x ) is measured in mms. The time period and the amplitude of the oscillations are A. The amplitude is ( 5 mathrm{cms} ) and time period is 1 sec B. The amplitude is 5 ( mathrm{mms} ) and time period is ( 1 mathrm{sec} ) c. The amplitude is ( 5 mathrm{mms} ) and time period is 3.142 seo D. The amplitude is ( 5 mathrm{cms} ) and time period is 3.142 sec | 11 |
| 121 | The variation of PE of a simple harmonic oscillator is as shown. Then force constant of the system is (PE ‘U’ is in joules, displacement ( ^{prime} x^{prime} ) is in ( mathrm{mm} ) ) A. ( 100 mathrm{N} / mathrm{m} ) B. ( 150 mathrm{N} / mathrm{m} ) c. ( frac{200}{3} mathrm{N} / mathrm{m} ) D. ( 300 mathrm{N} / mathrm{m} ) | 11 |
| 122 | Two pendulums differ in lengths by ( 22 mathrm{cm} . ) They oscillate at the same place so that one of them makes 30 oscillations and the other makes 36 oscillations during the same time. The lengths (in ( mathrm{cm} ) ) of the pendulums are ( mathbf{A} cdot 72 ) and 50 B. 60 and 38 c. 50 and 28 D. 80 and 58 | 11 |
| 123 | A spring is stretched by ( 0.20 m ) when a mass of ( 0.50 mathrm{kg} ) is suspended. A mass of ( 0.25 mathrm{kg} ) is suspended, then its period of oscillation will be ( left(g=10 m / s^{2}right) ) (approximately) A ( .0 .628 s ) B. ( 0.251 s ) ( mathbf{c} cdot 6.28 s ) D. 2.51s | 11 |
| 124 | An oscillator consists of a block attached to spring ( (mathrm{K}=1,1,400 mathrm{l}, mathrm{N} / mathrm{m}), mathrm{At} ) some time ( t, ) the position (measured from the system’s equilibrium location), velocity and acceleration of the block are ( boldsymbol{x}=mathbf{0 . 1 0 0 m}, boldsymbol{v}= ) ( -15.0 m / s, ) and ( a=-90 m / s^{2} . ) The amplitude of the motion and the mass of the block are : A. ( 0.2 m, 0.84 k g ) ( g ) в. ( 0.3 m, 0.76 k g ) ( mathrm{c} .0 .4 m, 0.54 mathrm{kg} ) D. ( 0.5 m, 0.44 k g ) | 11 |
| 125 | A glider is oscillating in SHM on an air track with an amplitude A. You slow it so that its amplitude becomes half. Find the total mechanical energy in terms of previous value. A ( cdot 1 / 2 ) в. ( 1 / 3 ) c. ( 1 / sqrt{5} ) D. ( 1 / 4 ) | 11 |
| 126 | A particle of mass ( m ) is oscillating with amplitude ( A ) and angular frequency ( omega ) its average energy in one time period is ? A ( cdot frac{1}{2} m omega^{2} A^{2} ) B ( cdot frac{1}{4} m omega^{2} A^{2} ) ( mathrm{c} cdot m omega^{2} A^{2} ) D. zero | 11 |
| 127 | Which of the following quantities is always negative in SHM? This question has multiple correct options ( mathbf{A} cdot vec{F} cdot vec{a} ) в. ( vec{F} cdot vec{r} ) c. ( vec{v} . vec{r} ) D. vecू.न | 11 |
| 128 | In SHM the net force towards mean position is related to its displacement (x) from mean position by the relation ( mathbf{A} cdot F propto x ) B . ( F propto frac{1}{x} ) c. ( F propto x^{2} ) D. ( F propto frac{1}{x^{2}} ) | 11 |
| 129 | A string of length ( 0.5 m ) carries a bob with a period ( 2 pi s . ) Calculate the angle of inclination of string with vertical and tension in the string. | 11 |
| 130 | The equation of a simple harmonic motion is ( boldsymbol{x}=mathbf{0 . 3 4} cos (mathbf{3 0 0 0 t}+mathbf{0 . 7 4}) ) where ( x ) and ( t ) are in ( mathrm{mm} ) and ( mathrm{sec} ) respectively. The frequency of the motion is : A. 3000 B. 3000/2pi c. ( 0.74 / 2 pi ) D. 3000/ ( pi ) | 11 |
| 131 | The displacement of a particle is represented by the equation ( y=sin ^{3} omega t ) the motion is A. Non-periodic B. Simple harmonic with period ( pi / omega ) c. simple harmonic with period ( 2 pi / omega ) D. Periodic but not simple harmonic | 11 |
| 132 | Vertical displacement of a plank with a body of mass ‘m’ on it is varying according to law ( y=sin omega t+sqrt{3} cos omega t ) The minimum value of ( omega ) for which the mass just breaks off the plank and the moment it occurs first after ( t=0 ) are given by: ( y is positive vertically upwards) A ( cdot sqrt{frac{g}{2}}, frac{sqrt{2}}{6} frac{pi}{sqrt{g}} ) В ( cdot frac{g}{sqrt{2}}, frac{2}{3} sqrt{frac{pi}{g}} ) c. ( sqrt{frac{g}{2}}, frac{pi}{3} sqrt{frac{2}{g}} ) D ( cdot sqrt{2 g}, sqrt{frac{2 pi}{3 g}} ) | 11 |
| 133 | The period of a particle in SHM is 8 seconds. At ( t=0 ) it is at the mean position. The ratio of the distance traveled by it in the 1 st and the 2nd seconds is A ( cdot frac{1}{2} ) B. ( frac{1}{sqrt{2}} ) ( c cdot sqrt{2} ) D. ( sqrt{2}+1 ) | 11 |
| 134 | In simple harmonic motion, loss of kinetic energy is proportional to ( mathbf{A} cdot e^{x}^{x} ) B ( cdot x^{3} ) ( c cdot log x ) D. ( x^{2} ) | 11 |
| 135 | ( frac{frac{2}{2}}{frac{1}{2}} ) | 11 |
| 136 | Mur man indower | 11 |
| 137 | The amplitude and time period of simple harmonic oscillator are ( a ) and ( T ) respectively. The time taken by it in displacing from ( boldsymbol{x}=mathbf{0} ) to ( boldsymbol{x}=boldsymbol{a} / mathbf{2} ) will be A . ( T ) в. ( frac{T}{2} ) c. ( frac{T}{4} ) D. ( frac{T}{12} ) | 11 |
| 138 | Which of the following quantities connected with SHM do not vary periodically? A. Total energy B. Velocity c. Displacement D. Acceleration | 11 |
| 139 | For a mass on a spring, which is maximized when the displacement of the mass from its equilibrium position is zero? A. Frequency B. Amplitude c. Period D. wavelength E. Kinetic Energy | 11 |
| 140 | A particle is performing S.H.M. with acceleration ( a=8 pi^{2}-4 pi^{2} x ) where ( x ) is coordinate of the particle w.r.t. the origin. The parameters are in S.I. units. The particle is at ( x=-2 a t t=0 ) then find the coordinate of the particle ( boldsymbol{w} . boldsymbol{r} . boldsymbol{t} ) origin at any time ( t ) | 11 |
| 141 | The displacement-time graph of a particle executing SHM is shown in the figure. Then This question has multiple correct options A. the velocity is maximum at ( t=T / 2 ) B. the acceleration is maximum at ( t=T ) C. the force is zero at ( t=3 T / 4 ) D. the potential energy equals the oscillation at t=T/2 | 11 |
| 142 | Figures show a sinusoidal wave at a given instant Which points are in phase? 4.4 .8 B. В, ( c cdot B, D ) D. c. | 11 |
| 143 | A particle is executing SHM of period 4s.Then the time taken by it to move from thr extreme position to ( frac{sqrt{3}}{2} ) of the amplitude is: A ( cdot frac{1}{3} s ) в. ( frac{2}{3} ) s c. ( frac{3}{4} ) D. ( frac{4}{3} ) s | 11 |
| 144 | Frequency of variation of kinetic energy of a simple harmonic motion of frequency n is A . ( 2 n ) в. ( n ) ( c cdot frac{n}{2} ) D. 3 n | 11 |
| 145 | The frequency of a particle executing SHM is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation, the spring is unstretched. Maximum speed of the particle is ( (boldsymbol{g}=mathbf{1 0 m} / boldsymbol{s} mathbf{2}) ) ( ^{text {A }} cdot frac{1}{2 pi} m s ) в. ( frac{7}{2 pi} m s ) c. ( frac{5}{2 pi} m s ) D. ( frac{1}{3 pi} m s ) | 11 |
| 146 | In an SHM, the total energy of a particle is A . a constant B. increases with time c. decreases with time D. oscillates with time | 11 |
| 147 | A certain weight is attached to a spring It is pulled down and then released. It oscillates up and down. Its K.E. will be. A. maximum at the middle of the movement B. maximum at the bottom c. maximum just before it is released D. constant | 11 |
| 148 | Time period of oscillation for given combination will be A ( cdot 2 pi sqrt{frac{mleft(K_{1}+K_{2}right)}{K_{1} K_{2}}} ) в. ( 2 pi sqrt{frac{m}{K_{1}+K_{2}}} ) c. ( 2 pi sqrt{frac{m K_{1} K_{2}}{K_{1}+K_{2}}} ) D. ( 2 pi sqrt{frac{m K_{1}}{K_{2}}} ) | 11 |
| 149 | Identify which of the following system exhibit simple harmonic motion? I. A simple pendulum with small deflection. II. A mass attached to a spring III. A ball bouncing up and down, in the absence of friction A. I only B. II only c. III only D. I and II only E . I, II, and III | 11 |
| 150 | For periodic motion of small amplitude ( A, ) the time period ( T ) of this particle is proportional to ( mathbf{A} cdot A sqrt{frac{m}{alpha}} ) B. ( frac{1}{A} sqrt{frac{m}{alpha}} ) c. ( A sqrt{frac{alpha}{m}} ) D. ( A sqrt{frac{2 alpha}{m}} ) | 11 |
| 151 | What type of curve do we get, if ( x^{2} ) and ( v^{2} ) are plotted for a particle executing ( mathrm{SHM}, mathrm{x} ) and ( mathrm{v} ) are the position and velocity of the particle: A . Circle B. Straight line c. Ellipse D. Rectangle | 11 |
| 152 | For a particle executing SHM, which kinetic parameter will be equal to zero At equilibrium position, A. velocity B. acceleration c. kinetic energy D. none of the above | 11 |
| 153 | The oscillations represented by curve 1 in the graph are expressed by equation ( x= ) Asinwt. The equation for the oscillations represented by curve 2 is expressed as: A. ( x=2 A sin (omega t-pi / 2) ) B. ( x=2 A sin (omega t+pi / 2) ) c. ( x=-2 operatorname{Asin}(omega t-pi / 2) ) D. ( x=operatorname{Asin}(omega t-pi / 2) ) | 11 |
| 154 | For a particle executing SHM. The KE ‘K’ is given by ( K=K_{0} cos ^{2} w t . ) The maximum value of PE is A. ( K_{0} ) B. zero c. ( frac{K_{0}}{2} ) D. ( frac{K_{0}}{4} ) | 11 |
| 155 | A particle executing simple harmonic motion with an amplitude A. The distance travelled by the particle in one time period is A. zero B. A c. 2 A D. 4A | 11 |
| 156 | An equilateral triangle has been constructed with an uniform wire whose resistance per unit length is 4 ohm ( c m^{-1} . ) If the length of each side of the triangle is ( 10 c, ) the resistance across any side will be A. 80 ohm B. 40 ohm c. ( frac{80}{3} ) onm D. ( frac{40}{3} ) ohm | 11 |
| 157 | Force acting on a block is ( boldsymbol{F}=(-4 x+ ) 8). Here ( F ) is in Newton and ( x ) the position of block on ( x ) -axis in meters A. motion of the block is periodic but not simple harmonic B. motion of the block is not periodic c. motion of the block is simple harmonic about origin, ( x=0 ) D. motion of the block is simple harmonic about ( x=2 m ). | 11 |
| 158 | Column 1 a) A linear S.H.M e) ( frac{d^{2} Theta}{d t^{2}}=frac{c}{Theta} ) b) Angular S.H.M f) ( frac{d^{2} x}{d t^{2}}+ ) ( frac{boldsymbol{R}}{boldsymbol{m}} frac{boldsymbol{d} boldsymbol{x}}{boldsymbol{M} boldsymbol{t}}+boldsymbol{x} boldsymbol{w}^{2}=frac{boldsymbol{F}}{boldsymbol{m}} cos boldsymbol{theta} ) c) Damped harmonic motion g) ( frac{d^{2} x}{d t^{2}}- ) ( frac{k}{m} x=0 ) d) forced oscillation h) ( m frac{d^{2} x}{d t^{2}}+ ) ( R frac{d n}{d t}+m x omega^{2}=0 ) A. a-e, ( b-h, c-g, d-f ) B. ( a-f, b-g, c-e, d-h ) ( c cdot a-g, b-e, c-h, d-f ) D. a-g, b-h, c-e, d-f | 11 |
| 159 | If the displacement of simple pendulum at any time is ( 0.02 m ) and acceleration is ( 2 m / s^{2}, ) then in this time angular velocity will be: ( mathbf{A} cdot 100 mathrm{rad} / mathrm{s} ) B. 10 rad/s c. 1 rad/s D. ( 0.1 mathrm{rad} / mathrm{s} ) | 11 |
| 160 | The equation of a simple harmonic motion is given by ( boldsymbol{x}=mathbf{6} sin mathbf{1 0 t}+ ) ( 8 cos 10 t, ) where ( x ) is in ( mathrm{cm}, ) and ( t ) is in seconds. Find the resultant amplitude A . 20 B . 10 c. 30 D. 50 | 11 |
| 161 | The potential energy function for a particle executing linear simple harmonic motion is given by ( U(x)= ) ( frac{1}{2} k x^{2}, ) where ( k ) is the force constant. For ( k=0.5 N m^{-1}, ) the ( operatorname{graph} ) of ( U(x) ) versus ( x ) is shown in figure. Show that a particle of total energy ( 1 J ) moving under this potential’ turns back’ when it reaches ( boldsymbol{x}=pm mathbf{2} boldsymbol{m} ) | 11 |
| 162 | A sings with a frequency ( (n) ) and ( B ) sings with a frequency ( 1 / 8 ) that of ( A ). If the energy remains the same and the amplitude of ( A ) is ( a ), then amplitude of ( boldsymbol{B} ) will be A . ( 2 a ) B. ( 8 a ) c. ( 4 a ) D. | 11 |
| 163 | The force on a particle of mass ( 10 g ) is ( (10 hat{i}+5 hat{j}) N . ) If it starts from rest, what would be its position at time ( t=5 s ? ) A . ( (12500 hat{i}+6250 hat{j}) m ) B. ( (6250 hat{imath}+12500 hat{j}) m ) c. ( (12500 hat{i}+12500 hat{j}) m ) D. ( (6250 hat{i}+6250 hat{j}) m ) | 11 |
| 164 | The differential equation representing the SHM of a particuleis ( frac{mathbf{9 d}^{2} boldsymbol{y}}{boldsymbol{d t}^{2}}+mathbf{4} boldsymbol{y}=mathbf{0} ) The time period of the particle is given by A ( cdot frac{pi}{3} sec ) в. ( pi ) sec ( ^{mathrm{c}} cdot frac{2}{pi} 3 s e c ) D. ( 3 pi s e c ) | 11 |
| 165 | The time taken by the particle in SHM for maximum displacement is ? A. ( T / 8 ) B. T/6 ( c cdot pi / 2 ) D. T/4 | 11 |
| 166 | The diagram in Fig. shows the displacement- time graph of a vibrating body. Name the kind of vibrations. | 11 |
| 167 | Assertion (A): The displacement time graph for a particle in SHM begins from mean position. Reason (R): The displacement of a particle in SHM is given by ( y=A sin omega t ) A. Both A and R are true and R is the correct explanation of A B. Both A and R are true and R is not the correct explanation of C. A is true and R is false D. A is false and R is true | 11 |
| 168 | The displacement of a particle is represented by the equation ( y= ) ( sin ^{3}(omega t) . ) The motion is A. non-periodic B. periodic but not simple harmonic c. simple harmonic with period ( frac{2 pi}{omega} ) D. simple harmonic with period ( frac{pi}{omega} ) | 11 |
| 169 | The average acceleration in one time period in a simple harmonic motion is A ( cdot A omega^{2} ) B. ( A omega^{2} / 2 ) C. ( A omega^{2} / sqrt{2} ) D. zero | 11 |
| 170 | Figure gives the ( x ) -t plot of particle executing one-dimensional simple harmonic motion.Give the signs of position, velocity and acceleration variables of the particle at ( t=0.3 s, 1.2 s ) -1.2 s. | 11 |
| 171 | Kinetic energy of the bob of a simple pendulum is maximum at the extreme left position. State whether true or false. A. True B. False | 11 |
| 172 | What is the relation between restoring force ( (boldsymbol{F}) ) and displacement ( (boldsymbol{x}) ) of ( mathbf{a} ) particle to perform SHM. A ( . F propto x^{2} ) B. ( F propto x^{-2} ) c. ( F propto x^{1 / 2} ) D. ( F propto x ) | 11 |
| 173 | A mass of ( 5 mathrm{kg} ) is suspended on a spring of stiffness ( 4000 mathrm{N} / mathrm{m} ). The system is fitted with a damper with a damping ratio of 0.2. The mass is pulled down 50 ( mathrm{mm} ) and released. Calculate the displacement after 0.3 sec: ( mathbf{A} cdot 4.07 mathrm{mm} ) в. ( 4.7 mathrm{mm} ) ( c .7 .4 mathrm{mm} ) D. 7.04 mm | 11 |
| 174 | A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road. If the acceleration is ( a_{0} ) and the length of the pendulum is ( l ), find the time period of small oscillations about the mean position ( mathbf{A} ) B. ( ^{mathrm{C}} 2 pi frac{sqrt{l}}{left(a_{0}^{2}+g^{2}right)^{1 / 4}} ) D. ( pi frac{sqrt{l}}{left(a_{0}^{2}+g^{2}right)^{1 / 4}} ) | 11 |
| 175 | If a particle is executing ( S H M ) on a straight line. ( A ) and ( B ) are two points at which its velocity is zero. It passes through a certain point ( boldsymbol{P}(boldsymbol{A} boldsymbol{P}<boldsymbol{P B}) ) at successive intervals of 0.5 sec and 1.5 sec with a speed of ( 3 mathrm{m} / mathrm{s} ). The maximum speed of the particle is A. ( 3 m / s ) B. ( 3 sqrt{2} mathrm{m} / mathrm{s} ) ( mathbf{c} cdot 3 sqrt{3} m / s ) D. ( 6 m / s ) | 11 |
| 176 | The propagation of a sound wave in a gas is given by the equation ( frac{boldsymbol{d}^{2} boldsymbol{y}}{boldsymbol{d} boldsymbol{t}^{2}}=frac{boldsymbol{k}}{boldsymbol{l}} cdot frac{boldsymbol{d}^{2} boldsymbol{y}}{boldsymbol{d} boldsymbol{x}^{2}} ) Then the velocity of sound is given by A ( cdot frac{k}{l} ) в. ( frac{l}{k} ) c. ( sqrt{left(frac{k}{l}right)} ) D. ( sqrt{left(frac{l}{k}right)} ) | 11 |
| 177 | A particle executes SHM with a period of ( T ) seconds. ( t ) seconds after it has crossed the equilibrium position it is at a point ( P . ) After how much time it will be again at ( P ? ) A ( cdot frac{T}{2}-t ) в. ( frac{T}{4}-t ) c. ( frac{T}{2}-2 t ) D. ( frac{T}{4}-2 t ) | 11 |
| 178 | Which of the following equations does not represent a simple harmonic motion A. ( y-alpha omega t ) B. ( y=alpha omega t ) c. ( y=alpha omega t+b c o s omega t ) D. ( y=alpha ) tanct | 11 |
| 179 | The amplitude of oscillation of a particle is 0.05 m.lf its period is 1.57 s. Then the velocity at the mean position is ( A cdot 0.1 mathrm{m} / mathrm{s} ) B. ( 0.2 mathrm{m} / mathrm{s} ) ( c cdot 0.3 mathrm{m} / mathrm{s} ) D. ( 0.5 mathrm{m} / mathrm{s} ) | 11 |
| 180 | An object performing S.H.M. with mass of ( 0.5 k g, ) force constant ( 10 N / m ) and amplitude ( 3 c m . ) What is the speed at ( x=2 ? ) | 11 |
| 181 | The frequency of oscillation in an oscillation is A ( cdot frac{pi}{omega} ) в. ( frac{2 pi}{omega} ) c. ( frac{omega}{2 pi} ) D. ( frac{omega}{pi} ) | 11 |
| 182 | The motion of a particle executing SHM is given by ( boldsymbol{x}=mathbf{0 . 0 1} sin 100 pi(boldsymbol{t}+mathbf{0 . 0 5}) ) where ( x ) is in metre and ( t ) in second. The time period of motion(in second) is: A . 0.01 B. 0.02 c. ( 0 . ) D. 0.2 | 11 |
| 183 | Which of the following is an example of oscillatory motion? A. Heart beat of a persion B. Motion of earth around the sun c. Motion of Hally’s comet around the sun D. oscillations of a simple pendulam | 11 |
| 184 | What are Damped vibrations? | 11 |
| 185 | The maximum displacement of a particle executing SHM from its mean position is ( 2 mathrm{cm} ) and its time period is 1 s. The equation of its displacement will be A . ( x=2 sin 4 pi t ) B. ( x=2 sin 2 pi t ) c. ( x=sin 2 pi t ) D. ( x=4 sin 2 pi t ) | 11 |
| 186 | A particle executing SHM according to the equation ( boldsymbol{x}=mathbf{5} cos left[mathbf{2} boldsymbol{pi} boldsymbol{t}+frac{pi}{4}right] ) in ( mathbf{S} ) units. The displacement and acceleration of the particle at ( t=1.5 ) s is: A ( .-3.0 m, 100 m s^{-2} ) B . ( +2.54 m, 200 m s^{-2} ) c. ( -3.54 m, 140 m s^{-2} ) D. ( -3.55 m, 120 m s^{-2} ) | 11 |
| 187 | ( sum_{k} ) | 11 |
| 188 | A particle executes a simple harmonic motion of time period ( T . ) Find the time taken by the particle to go directly from its mean position to half the amplitude. A ( cdot frac{T}{2} ) в. ( frac{T}{4} ) c. ( frac{T}{8} ) D. ( frac{T}{12} ) | 11 |
| 189 | Out of the following functions representing motion of particle which represents SHM? ( x=sin ^{3} omega t ) 2. ( x=1+omega t+omega^{2} t^{2} ) 3. ( boldsymbol{x}=cos omega boldsymbol{t}+cos boldsymbol{3} boldsymbol{omega} boldsymbol{t}+boldsymbol{c} boldsymbol{o} boldsymbol{s} boldsymbol{5} boldsymbol{t} ) 4. ( x=sin omega t+cos omega t ) A. Only 1 B. Only 1 and 3 c. only 1 and 4 D. only 4 | 11 |
| 190 | A rod of moment of inertia I and length is suspended from a fixed end and given small oscillations about the point of suspension, the restoring torque is found to be ( -(boldsymbol{m} boldsymbol{g} boldsymbol{L} / mathbf{2}) boldsymbol{s} boldsymbol{i} boldsymbol{n} boldsymbol{theta} . ) What will be the angular equation of motion of the SHM A ( cdot frac{d^{2} theta}{d t^{2}}+(m g L / I) theta=0 ) B ( cdot frac{d^{2} theta}{d t^{2}}+(m g L / 2 I) theta=0 ) ( ^{mathbf{c}} cdot frac{d^{2} theta}{d t^{2}}+(2 m g L / I) theta=0 ) D. ( frac{d^{2} theta}{d t^{2}}+(m g L / 4 I) theta=0 ) | 11 |
| 191 | Mass ( m_{1} ) hits and sticks with ( m_{2} ) while sliding horizontally with velocity ( v ) along the common line of centres of the three [ text { equal masses }left(boldsymbol{m}_{1}=boldsymbol{m}_{2}=boldsymbol{m}_{3}=boldsymbol{m}right) ] Initially masses ( m_{2} ) and ( m_{3} ) are stationary and the spring is unstretched. Minimum kinetic energy of ( m_{2} ) is ( y m v^{2} / 36 . ) Find ( y ) | 11 |
| 192 | Assertion A particle is moving along x-axis. The resultant force ( boldsymbol{F} ) acting on it at position ( x ) is given by ( F=-a x-b ) Where ( a ) and ( b ) are both positive constants. The motion of this particle is not SHM Reason In SHM, restoring force must be proportional to the displacement from mean position. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 11 |
| 193 | If ( x ) denotes displacement in time ( t ) and ( boldsymbol{x}=boldsymbol{a} cos omega boldsymbol{t}, ) then for ( boldsymbol{omega}=mathbf{1}, ) the acceleration will be ( mathbf{A} cdot a cos t ) B. – a cost ( mathbf{c} cdot a sin t ) D. – ( a sin t ) | 11 |
| 194 | Two particles A and B perform SHM along the same straight line with the same amplitude ‘a’, same frequency ‘f and same equilibrium position ‘0’. The greatest distance between them is found to be ( 3 a / 2 ). at some instant of time they have the same displacement from mean position. What is this displacement? A ( cdot frac{a}{sqrt{2}} ) B. ( a sqrt{7} / 4 ) c. ( sqrt{7} a / 4 ) D. 3a/4 | 11 |
| 195 | A particle undergoes SHM. When its displacement is ( 8 mathrm{cm}, ) it has speed 3 ( mathrm{m} / mathrm{s} ) and a speed ( 4 mathrm{m} / mathrm{s} ) at displacement of ( 6 mathrm{cm} . ) Find time period of oscillation. A . ( 4 pi ) в. ( 6 pi ) c. ( 7 pi ) D. ( 8 pi ) | 11 |
| 196 | Two blocks ( A ) and ( B ), each of mass ( m ), are connected by a massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in Fig. A third identical block ( C ), also of mass ( m ) moves on the floor with a speed ( mathbf{v} ) along the line joining ( A ) and ( B ) and collides with ( A, ) then This question has multiple correct options A. The KE of the AB system at maximum compression of the spring is zero. B. The KE of the AB system at maximum compression of the spring is ( (1 / 4) m v^{2} ) c. The maximum compression of the spring is ( v sqrt{frac{m}{k}} ) D. The maximum compression of the spring is ( v sqrt{frac{m}{2 k}} ) | 11 |
| 197 | A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of time t. Which of the following statements is true. Assume that the particle started at ( t=0 ) at its mean position A. P.E. is maximum when ( t=0 ) B. T.E. is zero when ( t=0 ) c. K.E. is maximum when ( t=0 ) D. K.E. is minimum when t = | 11 |
| 198 | The time period of a particle executing SHM is 2 s. After what time interval from ( t=0, ) will its displacement be half the amplitude: ( A cdot(1 / 4) sec s ) B. (1/3) secs c. (1/2) secs D. 1 sec | 11 |
| 199 | Two particles executing SHM of same frequency, meet at ( x=+A / 2, ) while moving in opposite directions. Phase difference between the particles is | 11 |
| 200 | A long spring when stretched by ( x c m ) has a potential energy ( v . ) On increasing the stretching to ( n x c m, ) the potential energy stored in the spring will be A ( . n v ) B . ( n^{2} v ) ( c cdot n v^{3} ) D. ( n v^{-2} ) | 11 |
| 201 | The force of a required to row a boat at velocity is proportional to square of its speed of ( mathbf{v} mathrm{km} / mathrm{h} ) requires ( 4 mathrm{KW}, ) how many does a seepd of ( 2 mathrm{V} mathrm{km} / mathrm{h} ) required ( A cdot 8 mathrm{kW} ) B. 16 kw c. з२кw D. 76kw | 11 |
| 202 | A particle is an oscillating simple harmonically with angular frequency ( omega ) and amplitude ( A ). It is at a point ( (A) ) at ( a ) certain instant (shown in the figure). At this instant, it is moving towards mean position (B). It takes time ( t ) to reach mean position (B). If the time period of oscillation is ( T, ) the average speed between ( A ) and ( B ) is : A. ( frac{A sin omega t}{t} ) B. ( frac{A cos omega t}{t} ) c. ( frac{A sin omega t}{T} ) D. ( A cos omega t ) | 11 |
| 203 | In reality, a spring won’t oscillate for ever. will ( quad ) the amplitude of oscillation until eventually the system is at rest. A. Frictional force, increase B. Viscous force, decrease c. Frictional force, decrease D. Viscous force, increase | 11 |
| 204 | A plank with a small block on top of it is undergoing vertical SHM. Its period is 2 sec.The minimum amplitude at which the block will separate from plank is : A ( cdot frac{10}{pi^{2}} ) в. ( frac{pi^{2}}{10} ) c. ( frac{20}{pi^{2}} ) D. ( frac{pi}{10} ) | 11 |
| 205 | A partical having mass 10 g oscillates according to the equation ( boldsymbol{x}= ) ( (2.0 c m) sin left[left(100 s^{-1}right) t+frac{6}{pi}right] . ) Find (a) the amplitude, the time period and the force constant (b) the position, the velocity and the acceleration at ( t=0 ) | 11 |
| 206 | The displacement equations of two simple harmonic oscillators are given by ( boldsymbol{x}_{1}=boldsymbol{A}_{1} cos boldsymbol{omega} boldsymbol{t} ; boldsymbol{x}_{2}=boldsymbol{A}_{2} sin left(boldsymbol{omega} boldsymbol{t}+frac{pi}{6}right) ) The phase difference between them is: ( A cdot 30 ) B. 60 ( c cdot 90 ) D. ( 120^{circ} ) | 11 |
| 207 | The relation between ( T ) and ( g ) by ( mathbf{A} cdot T propto g ) в. ( T propto g^{2} ) c ( cdot T^{2} propto g^{-1} ) D ( T propto frac{1}{g} ) | 11 |
| 208 | A particle is executing simple harmonic simple motion of amplitude ( 5 mathrm{cm} ) and period 6 s. How long will it take to move from one end of its path on one side of mean position to a position ( 2.5 mathrm{cm} ) on the same side of the mean position? A . ( 1 mathrm{s} ) B. 1.5 s ( c cdot 3 s ) D. 3.5 | 11 |
| 209 | For the damped oscillator shown in Fig the mass of the block is ( 200 g, k= ) ( 80 N m^{-1} ) and the damping constant ( b ) is ( 40 g s^{-1} ) Calculate. (a) The period of oscillation, (b) Time period for its amplitude of vibrations to drop to half of its initial value (c) The time for the mechanical energy to drop to half initial value. | 11 |
| 210 | The oscillations of a pendulum slow down due to: A. the force exerted by air and the force exerted by friction at the support B. the force exerted by air only C. the forces exerted by friction at the support D. they never slow down | 11 |
| 211 | A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ‘b’, the correct equivalence would be : A. ( L leftrightarrow m, C leftrightarrow k, R leftrightarrow b ) в. ( L leftrightarrow frac{1}{b}, C leftrightarrow frac{1}{m}, R leftrightarrow frac{1}{k} ) c. ( L leftrightarrow k, C leftrightarrow b, R leftrightarrow m ) D ( cdot L leftrightarrow m, C leftrightarrow frac{1}{k}, R leftrightarrow b ) | 11 |
| 212 | On the average a human heart is found to beat 75 times in a minute. Calculate its beat frequency of heart and period. | 11 |
| 213 | A particle executes ( boldsymbol{S H} boldsymbol{M} ) on a line ( 8 c m ) long. Its ( K . E ) and ( P . E ) will be equal when its distance from the mean position is: A. ( 4 mathrm{cm} ) B. 2 cm ( mathrm{c} cdot 2 sqrt{2} mathrm{cm} ) D. ( sqrt{2} ) cm | 11 |
| 214 | toppr Q Type your question_ frictionless surface. The block is stretched by ( 0.2 m ) and released from rest at ( t=0 . ) It then executes simple harmonic motion with angular frequency ( omega=frac{pi}{3} r a d / s ) Simultaneously at ( t=0, ) a small pebble is projected with speed ( boldsymbol{v} ) from point ( P ) at an angle of ( 45^{circ} ) as shown in the figure. Point ( P ) is a a horizontal distance of ( 10 mathrm{m} ) from ( mathrm{O} ). If the pebble hits the block at ( t=1 ) sec, the value of ( v ) is ( left(operatorname{take} g=10 m / s^{2}right): ) ( mathbf{A} cdot sqrt{50} m / s ) B. ( sqrt{51} mathrm{m} / mathrm{s} ) ( mathbf{c} cdot sqrt{52} m / s ) D. ( sqrt{53} mathrm{m} / mathrm{s} ) | 11 |
| 215 | Two pulses in a stretched string whose centers are initially ( 8 mathrm{cm} ) apart are moving towards each other as shown in figure. The speed of each pulse is 2 ( mathrm{cm} / mathrm{s} . ) After 2 seconds, the total energy of the pulses will be : A . zero B. purely kinetic c. purely potential D. partly kinetic and partly potential | 11 |
| 216 | A particle of mass in oscillates with simple harmonic motion between two points ( x_{1} ) and ( x_{2}, ) the equilibrium position being a Its potential energy is plotted on the graph. Which of the following curve represents the phenomenon? A (a) B. c. D. | 11 |
| 217 | The phase difference between the velocity and displacement of a particle executing SHM is A ( cdot pi / 2 ) radian B. ( pi ) radian ( c cdot 2 pi ) radian D. zero | 11 |
| 218 | A train is moving at ( 30 m s^{-1} ) in still air. The frequency of the locomotive whistle is ( 500 mathrm{Hz} ) and the speed of sound is 345 ( m s^{-1} . ) The apparent wavelengths of sound in-front of and behind the locomotive are respectively. A. ( 0.65 mathrm{m}, 0.73 mathrm{m} ) B. ( 0.63 mathrm{m}, 0.75 mathrm{m} ) c. ( 0.60 mathrm{m}, 0.85 mathrm{m} ) D. ( 0.60 mathrm{m}, 0.75 mathrm{m} ) | 11 |
| 219 | Figure shows three systems in which a block of mass m can execute S.H.M. What is ratio of frequency of oscillation? ( A cdot 2: 1: 4 ) B. 1: 2: 4 c. 4: 2: 1 D. 3: 2: | 11 |
| 220 | Define phase of ( boldsymbol{S} . boldsymbol{H} . boldsymbol{M} ) ? | 11 |
| 221 | The figure shows an assembly consisting of a number of pendulums of varying lengths. The driver pendulum is pulled aside and released so that it oscillates in a plane perpendicular to that of the diagram. It will be observed that : A. all pendulums oscillate with the frequency of the driver pendulum and have the same amplitude. B. pendulums oscillate with different frequencies but equal amplitude, the shortest pendulum oscillating with the highest frequency, c. all pendulums oscillate with the frequency of the driver pendulum; the pendulum with length equal to that of the driver has the greatest amplitude D. pendulums oscillate with different frequencies and rent amplitudes | 11 |
| 222 | During the oscillations of a simple pendulum, the string takes the same path as that of the bob, then: A. motion of the bob and string are in SHM B. motion of the bob is SHM and that of the string is s angular SHM c. motion of the bob and string are angular SHM D. None of the above | 11 |
| 223 | Displacement vs time curve for a particle SHM is as shown in the figure: Which of the following statements is correct? A. Phase of the oscillator is same at ( t=0 ) s and ( t=2 ) s B. Phase of the oscillator is same at ( t=2 ) s and ( t=5 ) s c. Phase of the oscillator is same at ( t=1 ) s and ( t=7 ) s D. Phase of the oscillator is same at ( t=1 ) s and ( t=5 ) s. | 11 |
| 224 | A particle (1) of mass m initially at A is at rest. The radius of circular path is 1 m. The particle (1), collides at bottom with an other particle (2) of same mass m. The distance travelled by the particle (2) before coming at rest is ( 4.1 .5 mathrm{m} ) B. 2.5 ( c .3 .5 mathrm{m} ) ( D ) | 11 |
| 225 | You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags ( 15 mathrm{cm} ) when the entire automobile is places on ¡t. Also the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant ( k ) and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports ( 750 mathrm{kg} ) | 11 |
| 226 | Sitar maestro Ravi Shankar is playing sitar on its strings, and you, as a physicist (unfortunately without musical ears!), observed the following oddities. I. The greater the length of a vibrating string, the smaller its frequency. II. The greater the tension in the string, the greater is the frequency. III. The heavier the mass of the string, the smaller the frequency. IV. The thinner the wire, the higher its frequency. The maestro signalled the following combination as correct one. A. II, III and IV B. ।, / I and IV c. ।, ॥ land III D. I, II, III and IV | 11 |
| 227 | If we wish to represent the equation for the position of the mass in terms of a differential equation, which one of these would be the most suitable? A ( cdot quad m frac{d^{2} x}{d t^{2}}+b frac{d x}{d t}+k x=0 ) в. ( quad_{m} frac{d^{2} x}{d t^{2}}-b frac{d x}{d t}+k x=0 ) c. ( _{m} frac{d^{2} x}{d t^{2}}+b frac{d x}{d t}-k x=0 ) D. ( _{m} frac{d^{2} x}{d t^{2}}-b frac{d x}{d t}-k x=0 ) | 11 |
| 228 | A particle performs SHM with a period ( boldsymbol{T} ) and amplitude ( a ). The mean velocity of particle over the time interval during which it travels a distance ( a / 2 ) from the extreme position is: ( ^{A} cdot frac{6 a}{T} ) в. ( frac{2 a}{T} ) ( c cdot frac{3 a}{T} ) D. None | 11 |
| 229 | A block of mass ( m ) containing a net positive charges ( q ) is placed on a smooth horizontal table which terminates in a ventricles wall as shown in figure ( (29 . E 2) . ) The distance of the block from the wall is ( d . ) A horizontal electric field ( E ) towards right is switched on. Assuming elastic collisions ( if any) find the time periods of the resulting oscillatory motion. Is it a simple harmonic motion? | 11 |
| 230 | A particle is executing SHM along the ( x ) axis given by ( x=A ) sinomegat. What is the mangnitude of the average acceleration of the partical between ( t=0 ) and ( (T / 4) s ) where ( T ) is the time period of oscillation | 11 |
| 231 | Assertion All small oscillations are simple harmonic in nature. Reason Oscillations of spring block system are always simple harmonic whether amplitude is small or large. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 11 |
| 232 | Assertion : If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes double. Reason : The total energy is directly proportional to the amplitude of vibration of the harmonic oscillator. A. If both assertion and reason are true and reason is the correct explanation of assertion. B. If both assertion and reason are true and reason is not the correct explanation of assertion. c. If assertion is true but reason is false D. If both assertion and reason are false | 11 |
| 233 | A particle is executing ( S H M ) and its velocity ( v ) is related to its position ( (x) ) as ( y^{2}+a x^{2}=b, ) where a and b are positive constants. The frequency of oscillation of particle is ( ^{A} cdot frac{1}{2 pi} sqrt{frac{b}{a}} ) B. ( frac{sqrt{a}}{2 pi} ) ( c cdot frac{sqrt{b}}{2 pi} ) D. ( frac{1}{2 pi} sqrt{frac{a}{b}} ) | 11 |
| 234 | A transverse wave is passing through a medium. The maximum speed of the vibrating particle occurs when the displacement of the particle from the mean position is A. zero B. half of the amplitude c. equal to the amplitude D. none of the above | 11 |
| 235 | A simple harmonic oscillator has an amplitude a and time period T. The time required by it to a travel from ( x=a ) to ( x= ) a/2 is A. ( T / 6 ) B. T / 4 ( c cdot T / 3 ) D. T/2 | 11 |
| 236 | For a particle performing linear S.H.M. Its average speed over one oscillation is ( (a=a m p l i t u d e text { of } S . H . M, n= ) frequency of oscillation) A . ( 2 a n ) B. 4an c. ( 6 a n ) D. ( 8 a n ) | 11 |
| 237 | A person wearing a wrist watch that keeps correct time at the equator goes to N-pole. His watch will A. Keep correct time B. gain time c. loose time D. cannot say | 11 |
| 238 | The equation of the displacement of two particles making SHM are represented by ( boldsymbol{y}_{1}=a sin (omega t+phi) & boldsymbol{y}_{2}=a cos (boldsymbol{omega} boldsymbol{t}) ) The phase difference of the velocities of the two particles is : ( mathbf{A} cdot frac{pi}{2}+phi ) B. ( -phi ) c. ( phi ) D. ( phi-frac{pi}{2} ) | 11 |
| 239 | The disk has a weight of ( 100 mathrm{N} ) and rolls without slipping on the horizontal surface as it oscillates about its equilibrium position. If the disk is displaced, by rolling it counterclockwise 0.4 rad, determine the equation which describes its oscillatory motion when it is released. A ( cdot theta=-0.2 cos (16.16 t) ) B ( cdot theta=0.2 cos (16.16 t) ) c. ( theta=-0.4 cos (16.16 t) ) | 11 |
| 240 | The time period of a seconds’ pendulum A. 1 second B. 2 seconds c. 3 seconds D. 4 seconds | 11 |
| 241 | A particle is in S.H.M of amplitude ( 2 mathrm{cm} ) At extreme position the force is ( 4 mathrm{N} ). At the point mid-way between mean and extreme position, the force is : A. 1 N B. 2N c. 3 N D. 4 N | 11 |
| 242 | A particle is an linear simple harmonic motion between two extreme point ( mathbf{A} ) and B. ( 10 mathrm{cm} ) apart(See figure below) If the direction from ( A ) to ( B ) is taken as positive direction, what are signs of displacement ( x, ) velocity ( V ) and acceleration a, when the particle is at ( mathbf{A} ? ) A ( . x=-v e, V=-v e, a=-v e ) в. ( x=+v e, V=0, a=-v e ) c. ( x=+v e, V=-v e, a=+v e ) D. ( x=-v e, V=0, a=+v e ) | 11 |
| 243 | Find time period of ( mathrm{SHM}=? ) | 11 |
| 244 | The quantity ( frac{b}{2 sqrt{m k}} ) is called A. critical coefficient B. damping coefficient C. friction coefficient D. None of these | 11 |
| 245 | Two particles are executing simple harmonic motion of the same amplitude ( A ) and frequency ( omega ) along the x-axis. Their mean position is separated by distance ( boldsymbol{X}_{mathbf{0}}left(boldsymbol{X}_{mathbf{0}}>boldsymbol{A}right) . ) If the maximum separation between them is ( left(X_{0}+Aright), ) the phase difference between their motion is ( mathbf{A} cdot pi / 2 ) в. ( pi / 3 ) c. ( pi / 4 ) D. ( pi / 6 ) | 11 |
| 246 | A particle executes SHM with a time period of 4 s. At what instant of time, will the displacement of the particle be equal to the velocity of the particle ( A cdot 4 s ) B. 1 s ( c cdot 8 s ) D. Displacement and velocity cannot be equal at any instant | 11 |
| 247 | toppr Q Type your question time period is proportional to ( sqrt{frac{boldsymbol{m}}{boldsymbol{k}}}, ) as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of ( x=0 ) in a way different from ( k x^{2} ) and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the ( x ) -axis. Its potential energy is ( V(x)=alpha x^{4}(alpha>0) ) for ( |x| ) near the origin and becomes a constant equal to ( V_{0} ) for ( |x| geq X_{0}(text { see figure }) ) The acceleration of this particle for ( |boldsymbol{x}|>boldsymbol{X}_{0} ) is? A. Proportional to ( V_{0} ) B. Proportional to ( frac{V_{0}}{m X_{0}} ) C. Proportional to ( sqrt{frac{V_{0}}{m X_{0}}} ) D. Zero | 11 |
| 248 | A simple harmonic oscillation has an amplitude ( A ) and time period ( T . ) The time required to travel from ( boldsymbol{x}=boldsymbol{A} ) to ( boldsymbol{x}=frac{boldsymbol{A}}{2} ) is : A ( cdot frac{T}{6} ) в. ( frac{T}{4} ) c. ( frac{T}{3} ) D. ( frac{T}{12} ) | 11 |
| 249 | For what phase difference between two SHMs will the amplitude of the resultant SHM be zero A . ( pi / 2 ) в. ( 2 pi ) ( c . pi ) D. 0 | 11 |
| 250 | A particle executes SHM with amplitude ( 0.5 mathrm{cm} ) and frequency ( 100 s^{-1} ) The maximum speed of the particle is ( (operatorname{in} m / s) ) ( A ) B. 0.5 c. ( 5 pi times 10^{-5} ) D. ( 100 pi ) | 11 |
| 251 | The phenomenon in which the amplitude of oscillation of a pendulum decreases gradually is called A. decay period of oscillation B. damping c. building up of oscillation D. maintained oscillation | 11 |
| 252 | The angular frequency of a spring block system is ( omega_{0} ) This system is suspended from the ceiling of anelevator moving downwards with a constant speed ( v_{0} ) The block is at rest relative to the elevator. Lift issuddenly stopped. Assuming the downwards as a positive direction, choose the wrong statement: A ( cdot ) The amplitude of the block is ( frac{v_{0}}{omega_{0}} ) B. The initial phase of the block is ( pi ) C . The equation of motion for the block is ( frac{v_{0}}{omega_{0}} sin omega_{0} mathrm{t} ) D. The maximum speed of the block is ( v_{0} ) | 11 |
| 253 | Calculate the period of a wave, which is having the wavelenght ( 17 ~ m ) and wave velocity ( 340 m / s ) | 11 |
| 254 | A particle performs SHM given by the equation ( boldsymbol{x}= )Asinwt. Where is the particle at ( t=3 T / 8 ) A. The particle is at a distance of ( A / sqrt{(} 2) ) moving away from the mean position to its extreme B. The particle is at a distance of ( A / sqrt{(} 2) ) moving towards the mean position to its extreme C. The particle is at a distance of ( A(1-1 / sqrt{(} 2)) ) moving away from the mean position to its extreme D. None of these | 11 |
| 255 | A solid sphere of mass ( 2 mathrm{kg} ) is rolling on a frictionless horizontal surface with velocity ( 6 m / s . ) It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring ( =mathbf{3 6} N / boldsymbol{m} ) A ( cdot sqrt{14 m} ) в. ( sqrt{2.8} m ) c. ( sqrt{1.4 m} ) D. ( sqrt{0.7} m ) | 11 |
| 256 | A particle oscillates simple harmonically with a period of 16 s. Two second after crossing the equilibrium position its velocity becomes ( 1 mathrm{m} / mathrm{s} ). The amplitude is A ( cdot frac{pi}{4} mathrm{m} ) B. ( frac{8 sqrt{2}}{pi} mathrm{m} ) c. ( frac{8}{pi} mathrm{m} ) D. ( frac{4 sqrt{2}}{pi} mathrm{m} ) | 11 |
| 257 | The time period of periodic motion is calculated using A. Number of metres covered in 1 sec B. Number of 90 degree turns covered in 1 sec C. Number of oscillations per sec D. Number of hairpin bends particle covers in 1 sec | 11 |
| 258 | The equation of displacement of a particle executing SHM is ( boldsymbol{x}= ) ( 0.40 cos (2000 t+18) . ) The frequency of the particle is ( mathbf{A} cdot 10^{3} mathrm{Hz} ) B. ( 20 mathrm{Hz} ) c. ( 2 times 10^{3} mathrm{Hz} ) D. ( frac{10^{3}}{pi} mathrm{Hz} ) | 11 |
| 259 | A solid ball of mass ( mathrm{m} ) is made to fall form a height ( mathrm{H} ) on a pan suspended through a spring of spring constant ( mathrm{K} ) as shown in figure. if the ball does not rebound and the pan is mass less, then amplitude of oscillation is ( A cdot frac{m g}{K} ) ( ^{mathrm{B}} cdot frac{m g}{K}left(1+frac{2 H K}{m g}right)^{1 / 2} ) ( ^{mathrm{c}} frac{m g}{K}+left(frac{2 H K}{m g}right)^{1 / 2} ) ( ^{mathrm{D}} cdot frac{m g}{K}left[1+left(1+frac{2 H K}{m g}right)^{1 / 2}right] ) | 11 |
| 260 | A particle performs SHM with amplitude ( 25 mathrm{cm} ) and period 3s. The minimum time required for it to move between two mean positions is : A . 0.6 s в. 0.5 s ( c cdot 0.4 mathrm{s} ) D. 0.2 s | 11 |
| 261 | A particle performs S.H.M of amplitude A along a straight line. When it is at a distance ( frac{sqrt{3}}{2} A ) from mean position, its kinetic energy gets increased by an amount ( frac{1}{2} m omega^{2} A^{2} ) due to an impulsive force. Then its new amplitude becomes. A. ( frac{sqrt{5}}{2} ) B. ( frac{sqrt{3}}{2} ) A c. ( sqrt{2} mathrm{A} ) D. ( sqrt{5} mathrm{A} ) | 11 |
| 262 | A particle is in SHM. Then the graph of its acceleration as a function of displacement is a A . circle B. hyperbola c. straight line with negative slope D. straight line with positive slope | 11 |
| 263 | Three masses of ( 500 g, 300 g ) and ( 100 g ) are suspended at the end of a spring as shown and are in equilibrium. When the ( 500 g ) mass is removed suddenly, the system oscillates with a period of 2 second. When the ( 300 g ) mass is also removed, it will oscillate with a period ( mathbf{A} cdot 2 sec ) B. 4 sec c. 8 sec D. 1 sec | 11 |
| 264 | The time period of a particle executing ( boldsymbol{S H} boldsymbol{M} ) is ( boldsymbol{8} boldsymbol{s} . ) At ( boldsymbol{t}=boldsymbol{0} ) it is at the mean position. The ratio of distance covered by the particle in ( 1^{s t} ) second to the ( 2^{n d} ) second is A ( cdotleft(frac{1}{sqrt{2}-1}right) ) B. ( sqrt{2} ) c. ( (sqrt{2}+1) ) D. ( frac{1}{sqrt{2}} ) | 11 |
| 265 | The maximum velocity of a particle, executing simple harmonic motion with an amplitude ( 7 m m, ) is ( 4.4 m / s . ) The period of oscillation is A. ( 100 s ) B. ( 0.01 s ) ( c cdot 10 s ) D. ( 0.1 s ) | 11 |
| 266 | For a horizontal rod of moment of inertia I if a constant force ( F ) acts on the rod at an angle ( theta ) with the horizontal, the torque equation for the rod will be (NOTE: The rod is lying on a floor horizontally) A ( cdot I d^{2} theta / d t^{2}=2 F L theta ) B . ( I d^{2} theta / d t^{2}=F L theta ) ( mathbf{c} cdot I d^{2} theta / d t^{2}=(F L / 2) ) D. None of the above | 11 |
| 267 | Motion of an oscillating liquid in a tube is A. periodic but not simple harmonic B. non-periodic C. simple harmonic and time period is independent of the density of the liquid. D. simple harmonic and time period is directly proportional to the density of the liquid | 11 |
| 268 | A simple harmonic motion has an amplitude ( A ) and time period ( T . ) Find the time required by it to travel directly from ( boldsymbol{x}=mathbf{0} ) to ( boldsymbol{x}=boldsymbol{A} / mathbf{2} ) | 11 |
| 269 | A mass of ( 30 mathrm{kg} ) is supported on a spring of stiffness ( 60000 mathrm{N} / mathrm{m} ). The system is damped and the damping ratio is ( 0.4 . ) The mass is raised ( 5 mathrm{mm} ) and then released. Calculate the damped frequency in Hz. A. 6.235 B. 6.352 c. 6.523 D. 6.325 | 11 |
| 270 | Two particle are oscillating along two close parallel straight lines side by side with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is: A . 0 в. ( 2 pi / 3 ) ( c ) D. ( pi / 6 ) | 11 |
| 271 | Which one of the following statements is true for the velocity v and the acceleration a of a particle executing simple harmonic motion? A. When v is maximum, a is zero B. When v vis zero, a is zero c. When v is maximum, a is maximum D. Value of a is zero, whatever may be the value of ( v ) | 11 |
| 272 | A particle is executing S.H.M. with amplitude A and Time period T. Time taken by the particle to reach from extreme position to ( frac{A}{2} ) ( A cdot frac{T}{6} ) B・亞 ( c cdot frac{T}{3} ) ( D cdot frac{T}{4} ) | 11 |
| 273 | The potential energy of a particle with displacement ( x ) is ( U(x) . ) The motion is simple harmonic. If k is a positive constant then A ( cdot U=k x ) B. ( U=k ) c. ( U=-k x^{2} / 2 ) ( mathbf{D} cdot U=k x^{2} ) | 11 |
| 274 | A particle has a rectilinear motion and the figure gives its displacement as a function of time. Which of the following statements is false with respect to the motion? A. Between O and A the velocity is positive and acceleration is negative B. Between A and B the velocity and acceleration are positive c. Between B and C the velocity is negative and acceleration is positive D. Between C and D the acceleration is positive | 11 |
| 275 | The displacement of a particle executing simple harmonic motion is given by ( boldsymbol{y}=boldsymbol{A}_{0}+boldsymbol{A} sin omega boldsymbol{t}+boldsymbol{B} cos boldsymbol{omega} boldsymbol{t} ) Then the amplitude of its oscillation is given by : ( mathbf{A} cdot A_{0}+sqrt{A^{2}+B^{2}} ) B. ( sqrt{A^{2}+B^{2}} ) c. ( sqrt{A_{0}^{2}+(A+B)^{2}} ) D. ( A+B ) | 11 |
| 276 | A particle is executing SHM along a straight line. Its velocities at distances ( x_{1} ) and ( x_{2} ) from the mean position are ( V_{1} ) and ( V_{2} ) respectively. Its time period is: A ( cdot 2 pi sqrt{frac{V_{1}^{2}-V_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}} ) B ( cdot 2 pi sqrt{frac{x^{2}+x_{2}^{2}}{sqrt{1}+V_{2}^{2}}} ) C ( cdot 2 pi sqrt{frac{x^{2}-x^{2}}{V_{2}^{2}-V_{1}^{2}}} ) D. ( 2 pi sqrt{frac{v^{2}+V_{2}^{2}}{x^{2}+x_{2}^{2}}} ) | 11 |
| 277 | When a body is undergoing undamped vibration, the physical quantity that remains constant is A. amplitude B. velocity c. acceleration D. phase | 11 |
| 278 | The force on a particle of mass ( 10 g ) is ( (10 hat{i}+5 hat{j}) N . ) If it starts from rest, what would be its position at time ( t=5 s ? ) A . ( (12500 hat{i}+6250 hat{j}) m ) B. ( (6250 hat{imath}+12500 hat{j}) m ) c. ( (12500 hat{i}+12500 hat{j}) m ) D. ( (6250 hat{i}+6250 hat{j}) m ) | 11 |
| 279 | In simple harmonic motion, the acceleration is inversely proportional to the displacement of the body from its mean position. A. True B. False c. Nither D. Either | 11 |
| 280 | A particle of mass ( 1 mathrm{kg} ) is moving in a S.H.M with an amplitude of ( 0.02 mathrm{m} ) and a frequency of ( 60 mathrm{Hz} ). The maximum force acting on the particle is : A ( .2 .88 times 10^{3} mathrm{N} ) B . ( 1.44 times 10^{3} ) N c. ( 5.67 times 10^{3} mathrm{N} ) D. ( 0.75 times 10^{3} mathrm{N} ) | 11 |
| 281 | A mass of ( 50 mathrm{kg} ) is suspended from a spring of stiffness ( 10 mathrm{kN} / mathrm{m} ). It is set oscillating and it is observed that two successive oscillations have amplitudes of ( 10 mathrm{mm} ) and ( 1 mathrm{mm} ) Determine the damping ratio. A. 0.315 B. 0.328 c. 0.344 D. 0.353 | 11 |
| 282 | n figure, ( boldsymbol{k}=mathbf{1 0 0} boldsymbol{N} / boldsymbol{m}, boldsymbol{M}= ) 1 kgand ( F=10 N . ) A sharp blow by some external agent imparts a speed of ( 2 m / s ) to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant. A . ( 0.25 J ) в. 1.5 ( c cdot 2 J ) D. 2.5 5 | 11 |
| 283 | A body of mass ( 36 g m ), moves with S.H. ( M ) of amplitude ( A=13 mathrm{cm} ) and period ( boldsymbol{T}=mathbf{1 2} ) sec. At a time ( boldsymbol{t}=mathbf{0}, ) the displacement is ( boldsymbol{x}=+mathbf{1 3} ) cm. The shortest time of passage from ( boldsymbol{x}= ) ( +6.5 mathrm{cm} ) to ( x=-6.5 ) is ? ( mathbf{A} cdot 4 sec ) B. 2 sec ( mathrm{c} cdot 6 mathrm{sec} ) D. 3 sec | 11 |
| 284 | Two equal charges ( q ) are kept fixed at ( a ) and ( +a ) along the ( x ) -axis. A particle of mass ( m ) and charge ( frac{q}{2} ) is brought to the origin and given a small displacement along the ( x ) -axis, then: A. the particle executes oscillatory motion B. the particle remains stationary c. the particle executes, SHM along x-axis D. the particle executes SHM along y-axis | 11 |
| 285 | simple harmonic motion, then correct graph for acceleration ( ^{prime} a^{prime} ) and corresponding linear velocity’ ( v^{prime} ) is : ( A ) B. ( c ) ( D ) | 11 |
| 286 | The pendulum bob has a speed of ( 3 mathrm{m} / mathrm{s} ) at its lowest position. The pendulum is ( 0.5 mathrm{m} ) long. The speed of the bob, when the bob makes an angle of ( 60^{circ} ) to the vertical will be ( left(boldsymbol{g}=mathbf{1 0 m} / boldsymbol{s}^{2}right) ) ( A cdot 1 mathrm{m} / mathrm{s} ) B. ( 1.5 mathrm{m} / mathrm{s} ) ( c cdot 3 m / s ) D. ( 2 mathrm{m} / mathrm{s} ) | 11 |
| 287 | Tension in the string in the mean position of an oscillating pendulum is: A. Same as tension when it is not oscillating B. Depend upon the time speedd c. Less than tension when it is not oscillating D. More than tension when it is not oscillating | 11 |
| 288 | A particle moves along the ( x ) -axis according to the equation ( boldsymbol{x}= ) 10 ( sin 3(t) ). The amplitudes and frequencies of component SHMs are A ( cdot ) amplitude ( frac{30}{4}, frac{10}{4} ; ) frequencies ( frac{3}{2}, frac{1}{2} ) B. amplitude ( frac{30}{4}, frac{10}{4} ; ) frequencies ( frac{1}{2}, frac{3}{2} ) C. amplitude 10,( 10 ; ) frequencies ( frac{1}{2}, frac{1}{2} ) D. amplitude ( frac{30}{4}, 10 ; ) frequencies ( frac{3}{2}, 2 ) | 11 |
| 289 | When a mass ( m ) is connected individually to two springs ( S_{1} ) and ( S_{2} ), the oscillation frequencies are ( boldsymbol{v}_{1} ) and ( boldsymbol{v}_{2} ) If the same mass is attached to the two springs as shown in figure, the oscillation frequency would be: A ( cdot v_{1}+v_{2} ) B. ( sqrt{v_{1}^{2}+v_{2}^{2}} ) ( mathbf{c} cdotleft[frac{1}{v_{1}}+frac{1}{v_{2}}right]^{-1} ) D. ( sqrt{v_{1}^{2}-v_{2}^{2}} ) | 11 |
| 290 | Two S.H.M.’s are represented by the relations ( boldsymbol{x}=mathbf{4} sin (mathbf{8 0} boldsymbol{t}+boldsymbol{pi} / mathbf{2}) ) and ( boldsymbol{y}= ) ( 2 cos (60 t+pi / 3) ) The ratio of their time periods is A . 2: B. 1: c. 4: 3 D. 3: | 11 |
| 291 | toppr Q Type your question energy KE and time t is correctly represented by 4 ( B ) ( c ) ( D ) | 11 |
| 292 | A ball resting on a tray attached to a spring is made to oscillate vertically, such that the fixed end of the spring is attached to the ground. If the spring is compressed and let go, the ball will execute SHM until A. the ball is in contact with the spring B. the ball occasionally loses contact with the spring c. the ball loses contact with the spring beyond the mean position D. the ball loses contact with the spring beyond the extreme position | 11 |
| 293 | Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path. | 11 |
| 294 | A particle of mass ( 10 g ) is undergoing SHM of amplitude ( 10 mathrm{cm} ) and period ( 0.1 s . ) The maximum value of force on particle is about A ( .5 .6 N ) B. 2.75 N c. ( 3.5 N ) D. ( 4 N ) | 11 |
| 295 | constant ( mathrm{k} ) are attached horizontally at the two ends of a uniform horizontally rod ( A B ) of length ( ell ) and mass ( m ). The rod is pivoted at its centre ‘O’ and can rotate freely in horizontal plane. The other ends of the two spring are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is: ( mathbf{A} cdot frac{1}{2 pi} sqrt{frac{6 k}{m}} ) B ( cdot frac{1}{2 pi} sqrt{frac{2 k}{m}} ) ( mathbf{C} cdot frac{1}{2 pi} sqrt{frac{k}{m}} ) D. ( frac{1}{2 pi} sqrt{frac{3 k}{m}} ) | 11 |
| 296 | A plank with a body of mass ( m ) placed on it starts moving straight up according to the law ( y=a(1-cos omega t) ) where ( y ) is the displacement from the initial position, ( omega=11 ) rad/s. Find the minimum amplitude of oscillation of the plank at which the body starts falling behind the plank. | 11 |
| 297 | What will be the displacement equation of the simple harmonic motion obtained by combining the motions? ( boldsymbol{x}_{1}=mathbf{2} sin boldsymbol{omega} boldsymbol{t}, quad boldsymbol{x}_{2}=boldsymbol{4} sin left(boldsymbol{omega} boldsymbol{t}+frac{boldsymbol{pi}}{boldsymbol{6}}right) ) and ( quad x_{3}=6 sin left(omega t+frac{pi}{3}right) ) A ( cdot x=10.25 sin (omega t+phi) ) B cdot ( x=10.25 sin (omega t-phi) ) C ( cdot x=11.25 sin (omega t+phi) ) D ( cdot x=11.25 sin (omega t-phi) ) | 11 |
| 298 | A solid cylinder of mass ( mathrm{m} ) is attached to a horizontal spring with force constant k. The cylinder can roll without slipping along the horizontal plane. (See the accompanying figure.) Show that the center of mass of the cylinder executes simple harmonic motion with a period ( boldsymbol{T}=boldsymbol{2} boldsymbol{pi} sqrt{frac{boldsymbol{3} boldsymbol{m}}{boldsymbol{2} boldsymbol{k}}}, ) if displaced from mean position. | 11 |
| 299 | A wave is measured to have a frequency of ( 60 H z ). If its wavelength is ( 24 c m ) determine how fast it is moving. ( mathbf{A} cdot 24 m / s ) в. ( 10 mathrm{m} / mathrm{s} ) c. ( 12 m / s ) D. ( 14 mathrm{m} / mathrm{s} ) | 11 |
| 300 | One end of a rod of length ( L ) is fixed to a point on the circumference of a wheel of radius R. The other end is sliding freely along a straight channel passing through the centre 0 of the wheel as shown in the figure. The wheel is rotating with a constant angular velocity ( omega ) about ( 0 . ) Taking ( T=frac{2 pi}{P omega} ) the motion of the rod is? A. Simple harmonic with a period of ( T ) B. Simple harmonic with a period of ( T / 2 ) c. Not simple harmonic but periodic with a period of ( T ) D. Not simple harmonic but periodic with a period of ( T / 2 ) | 11 |
| 301 | topp grapillation Which one of the following ( T= ) time peric : | 11 |
| 302 | What is the minimum frequency of sound waves needed to prepare emulsion from two immiscible liquids? | 11 |
| 303 | For a particle showing motion number under the force ( F=-5(x-2)^{2} ), the motion is A. Translatory B. Oscillatiory c. sнm D. All of these | 11 |
| 304 | Time period of oscillation of a spring is 12s on earth. What shall be the time period if it is taken to moon? A . ( 6 s ) B. 12s c. ( 36 s ) D. 72 s | 11 |
| 305 | A simple harmonic motion has an amplitude ( A ) and time period T. Find the time required by it to travel directly from ( boldsymbol{x}=frac{boldsymbol{A}}{sqrt{mathbf{2}}} ) to ( boldsymbol{x}=boldsymbol{A} ) | 11 |
| 306 | The time taken by a particle performing SHM to pass from point ( A ) to ( B ) where its velocities are same is 2 s. After another ( 2 s ) it returns to ( B ). The ratio of distance ( O B ) to its amplitude (where ( O ) is the mean position) is : A. ( 1: sqrt{2} ) B . ( (sqrt{2}-1): 1 ) c. 1: 2 D. ( 1: 2 sqrt{2} ) | 11 |
| 307 | The actual frequency is A. ( 5.645 mathrm{Hz} ) B. 7.685 Hz c. ( 2.965 mathrm{Hz} ) D. 3.785 Нz | 11 |
| 308 | ( frac{k}{i} ) | 11 |
| 309 | A particle oscillating under a force ( overline{boldsymbol{F}}= ) ( -k bar{x}-b bar{v} ) is a (k and b are constants) A. simple harmonic oscillator B. linear oscillator c. damped oscillator D. forced oscillator | 11 |
| 310 | A particle of mass ( 0.1 k g ) executes ( S H M ) under a force ( boldsymbol{F}=(-10 x) N . ) Speed of particle at mean position is ( 6 m / s ). Then amplitude of oscillations is A. ( 0.6 m ) B. ( 0.2 m ) ( c cdot 0.4 m ) D. ( 0.1 m ) | 11 |
| 311 | A particle is performing oscillations with acceleration ( a=8 pi^{2}-4 pi^{2} x ) where ( x ) is coordinate of the particle w.r.t. the origin. The parameters are in S.I. units. The particle is at rest at ( x=2 ) at ( t=0 . ) Find coordinates of the particle w.r.t origin at any time. A ( .4-2 cos 2 pi t ) B. ( 2-2 cos 2 pi t ) c. ( 4-4 cos 2 pi t ) D. ( 2+4 sin 2 pi t ) | 11 |
| 312 | Q Type your question Based on the graphs, how does the motion between the two objects differ? Frequency II. Angular Frequency III. Period IV. Amplitude V. Phase Constant A . I and II B. I, II, and III C. I and III ( D ) E. All of them (I – v ) | 11 |
| 313 | f a particle is executing SHM, with an amplitude ( A, ) the distance moved and the displacement of the body in a time equal to its period are ( A cdot 2 A, A ) B. ( 4 A, ), ( c cdot A, A ) D. ( 0,2 A ) | 11 |
| 314 | A bar magnet oscillates with a frequency of10 oscillations per minute. When another bar magnet is placed on its axis at a small distance, it oscillates at 14 oscillations per minute. Now, the second bar magnet is turned so that poles are instantaneous, keeping the location same. The new frequency of oscillation will be ( A cdot 2 ) vibrations/min B. 4 vibrations/min c. 10 vibrations/min D. 14 vibrations/min | 11 |
| 315 | A ball is dropped on a hard floor and no energy is being lost to the floor. The ball bounces back to the same position, from where it was thrown. Is the ball executing SHM A. Yes, only for short displacements B. Yes, for all displacements c. No, for any kind of displacements D. None of the above | 11 |
| 316 | The angular frequency of small oscillations of the system shown in the figure is A ( cdot sqrt{(K / 2 m)} ) B . ( sqrt{(2 K / m)} ) c. ( sqrt{(K / 4 m)} ) D. ( sqrt{(4 K / m)} ) | 11 |
| 317 | What is the phase difference between acceleration and velocity of a particle executing simple harmonic motion? | 11 |
| 318 | ( operatorname{In} operatorname{an} operatorname{SHM} x=operatorname{asin} omega t, ) the minimum 0.5 second what is the minimum time A . ( 0.3 mathrm{s} ) B. ( 0.4 mathrm{s} ) ( c cdot 0.2 s ) D. ( 0.1 mathrm{s} ) | 11 |
| 319 | Can a traingle have A ( cdot 55^{circ}, 55^{circ} ) and ( 80^{circ} ) B . ( 33^{circ}, 74^{circ} ) and ( 73^{circ} ) c. ( 85^{circ}, 95^{circ} ) and ( 22^{circ} ) D. ( 25^{circ}, 95^{circ} ) and ( 32^{circ} ) | 11 |
| 320 | U is the PE of an oscillating particle and Fis the force acting on it at a given instant. Which of the following is true? A ( cdot frac{U}{F}+x=0 ) B. ( frac{2 U}{F}+x=0 ) c. ( frac{F}{U}+x=0 ) D. ( frac{F}{2 U}+x=0 ) | 11 |
| 321 | The diagram in Fig. shows the displacement- time graph of a vibrating body. Why is the amplitude of vibrations gradually decreasing? | 11 |
| 322 | A block is executing SHM on a rough horizontal surface under the action of an external variable force.The force is plotted against the position ( x ) of the particle from the mean position | 11 |
| 323 | The average kinetic energy of a simple harmonic oscillator with respect to mean position will be: A ( frac{k a^{2}}{6} ) в. ( frac{k a^{2}}{4} ) c. ( frac{k a^{2}}{3} ) D. ( frac{k a^{2}}{2} ) | 11 |
| 324 | The time taken by a particle performing ( boldsymbol{S H} boldsymbol{M} ) to pass from point ( boldsymbol{A} ) to ( boldsymbol{B} ) where its velocities are same is 2 seconds.After another 2 seconds it returns to ( B ). The time period of oscillation is (in seconds) A ( .2 s ) B. 8 s ( c cdot 6 s ) D. ( 4 s ) | 11 |
| 325 | The time period of the hour hand of a watch is ( A cdot 24 h ) B. 12 ( h ) ( c cdot 1 h ) D. 1 min | 11 |
| 326 | A transverse wave is described by the equation ( boldsymbol{y}=boldsymbol{y}_{0} sin 2 pileft(boldsymbol{f} boldsymbol{t}-frac{boldsymbol{pi}}{boldsymbol{lambda}}right) . ) The maximum particle velocity is equal to four times the wave velocity if : A ( cdot lambda=frac{pi y_{0}}{4} ) B. ( lambda=frac{pi y_{0}}{2} ) c. ( lambda=pi y_{0} ) D. ( lambda=2 pi y_{0} ) | 11 |
| 327 | Which of the following functions represent a simple harmonic motion? ( mathbf{A} cdot sin omega t-cos omega t ) B. ( sin ^{2} omega t ) ( mathbf{c} cdot sin omega t+sin ^{2} omega t ) D. ( sin omega t-sin ^{2} omega t ) | 11 |
| 328 | Assertion: If a block is in SHM, and a new constant force acts in the direction of change, the mean position may change. Reason :In SHM only variable forces should act on the body, for example spring force. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct and Reason is not the correct explanation for Assertion c. Assertion is correct and Reason is incorrect D. Assertion is incorrect and Reason is correct | 11 |
| 329 | Assertion : Every periodic motion is not simple harmonic motion. Reason: The motion governed by the force law ( boldsymbol{F}=-boldsymbol{k} boldsymbol{x} ) is simple harmonic. A. If both assertion and reason are true and reason is the correct explanation of assertion. B. If both assertion and reason are true and reason is not the correct explanation of assertion. c. If assertion is true but reason is false. D. If both assertion and reason are false | 11 |
| 330 | Two SHMs are given by ( boldsymbol{Y}_{mathbf{1}}= ) ( A sin left(frac{pi}{2} t+phiright) ) and ( Y_{2}=B sin left(frac{2 pi t}{3} t+right. ) ( phi ). The phase difference between these two after ‘1’ sec is A . ( pi ) в. ( frac{pi}{2} ) c. ( frac{pi}{4} ) D. | 11 |
| 331 | A simple harmonic oscillator of angular frequency 2 rad ( s^{-1} ) is acted upon by an external force ( F=sin t N . ) If the oscillator is at rest in its equilibrium position at ( t=0, ) its position at later times is proportional to A ( cdot sin t+frac{1}{2} cos 2 t ) B cdot ( operatorname{cost}-frac{1}{2} sin 2 t ) ( mathrm{c} cdot operatorname{sint}-frac{1}{2} sin 2 t ) D. ( sin t+frac{1}{2} sin 2 t ) | 11 |
| 332 | A particle executes SHM with a time period of 10 secs. At what times will the velocity of the particle be maximum, if at ( t=0, ) the particle starts from ( x=-A ) B. (0,10) c. (5,10) D. (2.5,7.5) | 11 |
| 333 | Q Type your question performing SHM with an amplitude of 1 ( m ). When the cannon reaches ( x= ) ( frac{sqrt{3}}{2} m ) from the equilibrium moving along +x direction, a bullet of ( frac{1}{2} k g ) is suddenly fired from it with ( a ) velocity of ( 20 m / s ) relative to the ground. What will be the new amplitude of the cannon? ( ^{A} cdot frac{sqrt{3}}{2} ) B. ( frac{sqrt{5}}{2} ) c. ( frac{sqrt{7}}{2} ) D. ( frac{sqrt{11}}{2} ) | 11 |
| 334 | Average Kinetic energy in one time period in an SHM is (Where ( m, omega ) and ( y_{0} ) are mass of particle, angular velocity and maximum displacement respectively) ( ^{A} cdot frac{m omega^{2} y_{0}^{2}}{2} ) B . ( m omega^{2} y_{0}^{2} ) c. ( frac{m omega^{2} y_{0}^{2}}{3} ) D. ( frac{m omega^{2} y_{0}^{2}}{4} ) | 11 |
| 335 | Two particles move parallel to ( x- ) axis about the origin with same amplitude ( a^{prime} ) and frequency ( omega . ) At a certain instant they are found at a distance ( a / 3 ) from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two? A ( cdot cos ^{-1} frac{7}{9} ) B. ( cos ^{-1} frac{5}{9} ) ( ^{mathbf{C}} cdot cos ^{-1} frac{4}{9} ) D. ( cos ^{-1} frac{1}{9} ) | 11 |
| 336 | The time period of the variation of potential energy of a particle executing ( boldsymbol{S H} boldsymbol{M} ) with period ( boldsymbol{T} ) is ( mathbf{A} cdot T / 4 ) в. ( T ) c. ( 2 T ) D. ( T / 2 ) E . ( T / 3 ) | 11 |
| 337 | Two particles ( A ) and ( B ) are revolving on two circles with time periods 4 second and 6 second respectively. Time period of particle A with respect to B will be A. 6 seconds B. 12 seconds c. 18 seconds D. 24 seconds | 11 |
| 338 | A particle executing SHM of amplitude ( 4 c m ) and ( T=4 s . ) The time taken by it to move from positive extreme position to half the amplitude is A . ( 1 s ) B. ( frac{1}{3} s ) c. ( frac{2}{3} ) D. ( sqrt{frac{3}{2}} s ) | 11 |
| 339 | A particle starts performing simple harmonic motion. Its amplitude is ( A ). At one time its speed is half that of the maximum speed. At this moment the displacement is: A ( cdot frac{sqrt{2} A}{3} ) B. ( frac{sqrt{3} A}{2} ) c. ( frac{2 A}{sqrt{2}} ) D. ( frac{3 A}{sqrt{2}} ) | 11 |
| 340 | A body oscillates with SHM according to the equation ( boldsymbol{x}=mathbf{5 . 0} cos (2 pi t+boldsymbol{pi}) . ) At time ( t=1.5 s, ) its displacement, speed and acceleration respectively is: B . ( 5,0,-20 pi^{2} ) c. ( 2.5,+20 pi, 0 ) D. ( -5.0,+5 pi,-10 pi^{2} ) | 11 |
| 341 | The equation of motion of a body in S.H.M is ( boldsymbol{x}=mathbf{4} sin left(boldsymbol{pi} boldsymbol{t}+frac{pi}{3}right) . ) The frequency, per minute, of the motion is | 11 |
| 342 | The potential energy of a simple pendulum will be maximum when it is- A. At the turning points of oscillations B. At the equilibrium C. In between the above two cases D. At any position, it has always a fixed value | 11 |
| 343 | A string fixed at both ends, oscillate in ( 4^{t h} ) harmonic. The displacement of a particle of string is given as: ( boldsymbol{Y}=boldsymbol{2} boldsymbol{A} sin (boldsymbol{5} boldsymbol{pi} boldsymbol{x}) cos (boldsymbol{1} boldsymbol{0} boldsymbol{0} boldsymbol{pi} boldsymbol{t}) . ) Then find the length of the string? ( mathbf{A} cdot 80 mathrm{cm} ) B. 100 ст ( mathbf{c} .60 mathrm{cm} ) D. ( 120 mathrm{cm} ) | 11 |
| 344 | The displacement ( y ) of a executing Periodic motion is given by ( y=4 cos ^{2}left(frac{1}{2} tright) sin (1000 t) ) This expression may be considered to be a result of the superposition of independent harmonic motions. A. two B. three c. four D. five | 11 |
| 345 | A particle performs SHM with period and amplitude A. The mean velocity of the particle averaged over quarter oscillation, starting from right extreme position is ( mathbf{A} cdot mathbf{0} ) в. ( frac{2 A}{T} ) c. ( frac{4 A}{T} ) D. ( frac{3 A}{T} ) | 11 |
| 346 | A particle suspended from a fixed point, by a light inextensible thread of length is projected horizontally from its lowest position with velocity ( sqrt{frac{mathbf{7 g L}}{mathbf{2}}} . ) The thread will slack after swinging through an angle ( theta, ) such that ( theta ) equal A ( .30^{circ} ) B. ( 135^{circ} ) ( c cdot 120^{circ} ) D. ( 150^{circ} ) | 11 |
| 347 | The differential equation representing the SHM of a particule is ( frac{mathbf{9} d^{2} boldsymbol{y}}{boldsymbol{d} boldsymbol{t}^{2}}+boldsymbol{4} boldsymbol{Y}= ) 0.The time period of the particle is fiven by A ( frac{pi}{3} sec ) в. ( pi ) sec ( ^{mathrm{c}} cdot frac{2 pi}{3} s e c ) D. ( 3 pi s e c ) | 11 |
| 348 | Two blocks each of mass m is connected to the spring of spring constant ( mathrm{k} ) as shown in the figure. If the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. The time period of oscillation is A. ( 2 pi sqrt{frac{m}{k}} ) В ( .2 pi sqrt{frac{m}{2 k}} ) ( mathbf{c} cdot 2 pi sqrt{frac{m}{4 k}} ) D. ( 2 pi sqrt{frac{2 m}{k}} ) | 11 |
| 349 | When a particle executing SHM oscillates with a frequency’ ( omega^{prime}, ) then the kinetic energy of the particle A. Changes periodically with a frequency of ‘ ( omega^{prime} ) B. Changes periodically with a frequency of ‘2 ( omega ) ‘ C . changes periodically with a frequency of ( ^{prime} frac{omega}{2} ) D. Remains constant | 11 |
| 350 | The displacement of particle in SHM in one time period is A . zero B. a c. ( 2 a ) D. 42 | 11 |
| 351 | ILLUSTRATION 33.7 For different L (meter) values of L, we get different values of T. The curve between L v/s T is shown. Estimate g from this curve. (Take it = 10) 0.49 > 1452) | 11 |
| 352 | The displacement of the particle varies with time ( x=12 sin omega t-16 sin ^{3} omega t . ) If its motion is SHM, then maximum acceleration is A ( cdot 12 omega^{2} ) B. 36 ( omega^{2} ) c. ( 144 omega^{2} ) D. ( sqrt{192} omega^{2} ) | 11 |
| 353 | The maximum value attained by the tension in the string of a swinging pendulum is four times the minimum value it attains. There is no slack in the string. The angular amplitude of the pendulum is? A ( cdot 90^{circ} ) B. ( 60^{circ} ) ( c cdot 45^{circ} ) D. ( 30^{circ} ) | 11 |
| 354 | Assertion In ( S H M, ) acceleration is always directed towards the mean position. Reason ( ln S H M, ) the body has to stop momentary at the extreme position and move back to mean position. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 11 |
| 355 | The phase space diagram for simple Momentum harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initia conditions, and ( E_{1} ) and ( E_{2} ) are the total mechanical energies respectively. Then: ( mathbf{A} cdot E_{1}=sqrt{2} E_{2} ) ( mathbf{B} cdot E_{1}=2 E_{2} ) ( mathbf{c} cdot E_{1}=4 E_{2} ) ( mathbf{D} cdot E_{1}=16 E_{2} ) | 11 |
| 356 | A mass of ( 2.0 mathrm{kg} ) is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is ( 200 mathrm{N} / mathrm{m} ). What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take ( left.g=10 m / s^{2}right) ) A. ( 8.0 mathrm{cm} ) B. 10.0 ( mathrm{cm} ) c. Any values less than ( 12.0 mathrm{cm} ) D. 4.0 ( mathrm{cm} ) | 11 |
| 357 | A particle executes linear simple harmonic motion with an amplitude of ( 3 mathrm{cm} . ) When the particle is at ( 2 mathrm{cm} ) from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is ( ^{mathrm{A}} cdot frac{sqrt{5}}{pi} ) в. ( frac{sqrt{5}}{2 pi} ) c. ( frac{4 pi}{sqrt{5} pi} ) D. ( frac{2 pi}{sqrt{3}} ) | 11 |
| 358 | A particle executes simple harmonic motion between ( boldsymbol{x}=-boldsymbol{A} boldsymbol{a} boldsymbol{n} boldsymbol{d} boldsymbol{x}=+boldsymbol{A} ) The time taken for it to go from 0 to ( A / 2 ) to ( A ) is ( T_{2} ). Then A ( cdot T_{1}T_{2} ) c. ( T_{1}=T_{2} ) D. ( T_{1}=2 T_{2} ) | 11 |
| 359 | A uniform disc of mass ( M ) and radius ( R ) is suspended in vertical plane from a point on its periphery. Its time period of oscillation is A ( cdot 2 pi sqrt{frac{3 R}{g}} ) в. ( 2 pi sqrt{frac{R}{3 g}} ) ( ^{mathrm{c}} cdot_{2 pi} sqrt{frac{2 R}{3 g}} ) D. ( 2 pi sqrt{frac{3 R}{2 g}} ) | 11 |
| 360 | Velocity at mean position of a particle executing SHM is ( v ), then velocity of the particle at a distance equal to half of the amplitude is ( A cdot frac{sqrt{3}}{2} v ) B. ( frac{2}{sqrt{3}} v ) ( c cdot 2 v ) D. ( v ) | 11 |
| 361 | To show that a simple pendulum executing a simple harmonic motion it is necessary to assume that A. length of the pendulum is small. B. mass of the pendulum is small c. acceleration due to gravity is small. D. amplitude of the oscillation is small. | 11 |
| 362 | What to you understand by free vibrations of a body? | 11 |
| 363 | The angular frequency of the damped oscillator is given by ( omega=sqrt{left(frac{k}{m}-frac{r^{2}}{4 m^{2}}right)} ) where ( k ) is the spring constant, ( m ) is the mass of the oscillator and ( r ) is the damping constant. If the ratio ( frac{r^{2}}{m k} ) is ( 80 %, ) the change in time period compared to the undamped oscillator is approximately as follows: A. increases by 8% B. decreases by 8% C. increases by ( 1 % ) D. decreases by 1% | 11 |
| 364 | The time period of a particle executing SHM is ( 8 s . ) At ( t=0, ) it is at mean position. If ( A ) is the amplitude of particle, then the distance travelled by particle (in second) is? A ( cdot sqrt{2} A ) B ( cdotleft(1-frac{1}{sqrt{2}}right) ) c. ( 2 A ) D. ( frac{A}{sqrt{2}} ) | 11 |
| 365 | The pendulum of a bob undergoes SHM with an amplitude A is given by the equation ( boldsymbol{y}=boldsymbol{A} boldsymbol{s} boldsymbol{i n} boldsymbol{omega} boldsymbol{t} . ) The corresponding angular SHM equation can be written as A ( cdot theta=phi cos (omega t) ) в. ( theta=phi sin (omega t) ) c. ( theta=phi tan (omega t) ) D. ( theta=phi(omega t) ) | 11 |
| 366 | A particle is performing SHM of amplitude “A” and time period “T”. Find the time taken by the particle to go from 0 to ( A / sqrt{2} ) A . T/12 B. ( T / sqrt{2} ) c. ( pi / 6 ) D. ( t / 4 ) | 11 |
| 367 | A body executes S.H.M. under the action of a force ( boldsymbol{F}_{1} ) with a time period ( mathbf{4} / mathbf{5} ) seconds. If the force is changed to ( boldsymbol{F}_{2}, ) it executes S.H.M. with a time period ( 3 / 5 ) seconds. If both the forces ( F_{1} ) and ( F_{2} ) act simultaneously in the same direction on the body, then its time period in seconds is: A . ( 12 / 25 ) в. ( 24 / 25 ) c. ( 25 / 24 ) D. 25/12 | 11 |
| 368 | The velocity of a particle executing a simple harmonic motion is ( 13 m s^{-1} ) when its distance from the equilibrium position ( (Q) ) is ( 3 m ) and its velocity is ( 12 m s^{-1}, ) when it is ( 5 m ) away from ( Q ) The frequency of the simple harmonic motion is: ( mathbf{A} cdot frac{5 pi}{8} ) В. ( frac{5}{8 pi} ) c. ( frac{8 pi}{5} ) D. ( frac{8}{5 pi} ) | 11 |
| 369 | The acceleration (a) of SHM at mean position is : A. zero B. ( propto x ) ( c cdot propto x^{2} ) D. None of these | 11 |
| 370 | If the maximum velocity of a particle in SHM is ( v_{0}, ) then its velocity at half the amplitude from position of rest will be : A ( cdot v_{0} / 2 ) B. ( v_{0} ) c. ( v_{o} sqrt{3} / 2 ) D . ( v_{0} sqrt{3} / 4 ) | 11 |
| 371 | A simple harmonic motion has an amplitude ( A ) and time period ( T . ) The time required by it to travel ( x=A ) to ( boldsymbol{x}=frac{boldsymbol{A}}{2} ) is: A. ( T / 6 ) в. ( T / 4 ) c. ( T / 3 ) D. ( T / 2 ) | 11 |
| 372 | The equilibrium position of the particle is A. ( x=4 ) B. ( x=6 ) ( c cdot x=2 ) D. x = 3 | 11 |
| 373 | A particle executes SHM with a frequency f and its amplitude is A. If the frequency is doubled, the amplitude will be ( A cdot A / 2 ) B. A c. ( 2 A ) D. A/4 | 11 |
| 374 | The time period of a simple harmonic motion is ( 8 s . ) At ( t=0, ) it is at its equilibrium position. The ratio of distances transversed by it in the first and second seconds is: A ( cdot frac{1}{sqrt{2}} ) в. ( frac{1}{sqrt{2}-1} ) c. ( frac{1}{sqrt{3}} ) D. ( frac{1}{2} ) | 11 |
| 375 | In which of the following is the energy loss maximum? A. sonometer B. tuning fork c. thin tube D. broader pipe | 11 |
| 376 | A smooth inclined plane having angle of inclination ( 30^{circ} ) with horizontal has a mass 2.5 kg held by a sprang which is fixed at the upper end as shown in figure. If the mass is taken ( 2.5 mathrm{cm} ) up along the surface of the inclined plane, the tension in the spring reduces to zero If the mass is then released, the angular frequency of oscillation in radian per second is: A. 0.707 В. 7.07 c. 1.414 D. 14.14 | 11 |
| 377 | The period of oscillation of a simple pendulum of constant length is independent of ( A ). size of the bob B. shape of the bob c. mass of bob D. all of these | 11 |
| 378 | The amplitude of vibration of oscillating particle goes on decreasing because it is opposed by… A. resistive force of the medium B. resistive force by source oscillator c. resistive force by observer D. resistive force produced itself | 11 |
| 379 | The displacement ( x ) of a particle in motion is given in terms of time by ( x(x- ) 4) ( =1-5 cos omega t ) A. The particle executes SHM B. The particle executes oscillatory motion which is not SHM C. motion of the particle is neither oscillator; nor simple harmonic D. The particle is not acted upon by a force when it is at ( x ) ( =4 ) | 11 |
| 380 | Two masses ( m_{1} ) and ( m_{2} ) are connected to a spring of spring constant ( K ) at two ends. The spring is compressed by ( y ) and released. The distance moved by ( m_{1} ) before it comes to a stop for the first time is A. ( frac{m_{1} y}{m_{1}+m_{2}} ) в. ( frac{m_{2} y}{m_{1}+m_{2}} ) c. ( frac{2 m_{1} y}{m_{1}+m_{2}} ) D. ( frac{2 m_{2} y}{m_{1}+m_{2}} ) | 11 |
| 381 | A block tied between two identical springs is in equilibrium. If upper spring is cut, then the acceleration of the block just after the cut is ( 6 m s^{-2} ) Now if instead of upper string lower spring is cut, then the acceleration of the block just after the cut will be (Take ( left.g=10 m / s^{2}right) ) A ( .1 .25 mathrm{ms}^{-2} ) B. ( 4 m s^{-2} ) ( mathrm{c} cdot 10 mathrm{ms}^{-2} ) D. ( 2.5 mathrm{ms}^{-2} ) | 11 |
| 382 | Out of the following functions representing motion of a particle which represents ( boldsymbol{S H} boldsymbol{M} ) ? (A) ( y=sin omega t-cos omega t ) (B) ( y=sin ^{3} omega t ) (C) ( y=5 cos left(frac{3 pi}{4}-3 omega tright) ) (D) ( boldsymbol{y}=mathbf{1}+boldsymbol{omega} boldsymbol{t}+boldsymbol{omega}^{2} boldsymbol{t}^{2} ) A. Only ( (A) ) and ( (B) ) B. only ( (A) ) c. only ( (D) ) does not represent ( S H M ) D. only ( (A) ) and ( (C) ) | 11 |
| 383 | A student writes the equation of an angular ( mathrm{SHM} ) as ( boldsymbol{pi} / mathbf{3}=boldsymbol{pi} / mathbf{6} sin (boldsymbol{omega} boldsymbol{t}) . ) Is this correct? A. This equation is correct as it is of the form ( theta= ) ( phi sin (omega t) ) B. This equation is incorrect as instantaneous angular displacement is more than the angular amplitude C. This equation is incorrect as the equation lacks a phase factor D. This equation is correct since the SHM has a sin factor in it | 11 |
| 384 | Which of the following exhibit simple harmonic motion? I. A pendulum II. A mass attached to a spring III. A ball bouncing up and down, in the absence of friction A. I only B. II only c. ॥ only D. I and II only E . ।, ॥।, and III | 11 |
| 385 | A pendulum suspended from the ceiling of a train has a time period T when the train is at rest. When the train is accelerating with a uniform acceleration, the time period will ( A ). increase B. decrease c. become infinite D. remain unaffected | 11 |
| 386 | The graphs in figure show that a quantity y varies with displacement d in a system undergoing simple harmonic motion. The unbalanced force acting on the system. ( A ) B. I ( c ) ( D . ) iv | 11 |
| 387 | A 0.5 kg body performs simple harmonic motion with a frequency of 2 ( mathrm{Hz} ) and an amplitude of ( 8 mathrm{mm} ). Find the maximum velocity of the body, its maximum acceleration and the maximum restoring force to which the body is subjected. | 11 |
| 388 | The displacement of a particle executing simple harmonic motion is given by equation ( y=0.3 sin 20 pi(t+0.05) ) where time ( t ) is in second and displacement y is in metre. calculate the values of amplitude, time period initial phase and initial displacement of the particle | 11 |
| 389 | The amplitude and time period of a particle of mass ( 0.1 mathrm{kg} ) executing simple harmonic motion are ( 1 mathrm{m} ) and ( 6.28 mathrm{s} ) respectively. Then its (i)angular frequency, (ii) acceleration at a displacement of ( 0.5 mathrm{m} ) are respectively A ( cdot 1 mathrm{rad} / mathrm{s}, 0.5 mathrm{m} / mathrm{s}^{2} ) B. 2 rad/s, 1 m/s’ c. ( 0.5 mathrm{rad} / mathrm{s}, 0.5 mathrm{m} / mathrm{s}^{2} ) D. 1 rad/s, ( 1 mathrm{m} / mathrm{s}^{2} ) | 11 |
| 390 | A particle execute S.H.M from the mean position. its amplitude is ( A ), its time period is “T.’ At what displacement,its speed is half of its maximum speed. A ( cdot frac{sqrt{3} A}{2} ) в. ( frac{sqrt{2}}{3} A ) c. ( frac{2 A}{sqrt{3}} ) D. ( frac{3 A}{sqrt{A}} ) | 11 |
| 391 | The graph shows the variation in displacement with time for an object moving with simple harmonic motion. What is the maximum acceleration of the object? A . 10 B. 0.99 ( c .0 .44 ) D. none of the above | 11 |
| 392 | A particle is executing SHM with amplitude A and has maximum velocity ( V_{0} . ) Its speed at displacement A/2 will be | 11 |
| 393 | The displacement of a particle varies according to the relation ( boldsymbol{y}= ) ( 4(cos pi t+sin pi t) . ) The amplitude of the particle is: A. 8 units B. 2 units c. 4 units D. ( 4 sqrt{2} ) units | 11 |
| 394 | A body executing ( S . H . M . ) along a straight line has a velocity of ( 3 m s^{-1} ) when it is at a distance of 4 m from its mean position and ( 4 m s^{-1} ) when it is at a distance of ( 3 m ) from its mean position. Its angular frequency and amplitude are: A ( cdot 2 ) rad ( s^{-1} ) & ( 5 m ) B. 1 rad ( s^{-1} ) & ( 10 m ) c. 2 rad ( s^{-1} ) & ( 10 m ) D. 1 rad ( s^{-1} ) & ( 5 m ) | 11 |
| 395 | In SHM match the following Column I a) Maximum velocity e) ( 1 / 2 mathrm{M} ) ( omega^{2} A^{2} ) b) Maximum acceleration f) ( 1 / 4 mathrm{M} ) ( omega^{2} A^{2} ) c) Maximum Force g) ( A omega ) d) Maximum total energy h) ( omega^{2} A ) ¡) ( boldsymbol{m} boldsymbol{omega}^{2} boldsymbol{A} ) A. a-g, b-h, c-i, d-e B. a-h, b-g, c-i, d-e c. ( a-g, b-h, c-f, d-e ) D. a-g, b-i, c-h, d-f | 11 |
| 396 | A particle executes SHM about a point other than ( x=0 ) as shown in the graph Choose a CORRECT option(s) This question has multiple correct options A. Amplitude is equal to ( 4 m ) B. Equilibrium position is at ( x=0 ) C. Equilibrium position is at ( x=2 m ) Angular frequency ( frac{2 pi}{3} ) | 11 |
| 397 | A particle of mass ( mathrm{m} ) is performing simple harmonic motion along line AB with amplitude 2 a with centre of oscillation as ( 0 . ) At time ( t=0 ) particle is at point ( C(O C=a) ) and is moving towards B with velocity ( boldsymbol{v}=boldsymbol{a} sqrt{mathbf{3}} mathrm{m} / mathrm{s} ) The equation of motion can be given by This question has multiple correct options A. ( x=2 a(sin t+sqrt{3} cos t) ) B cdot ( x=2 a sin left(t+frac{pi}{6}right) ) c. ( x=a(sin t+sqrt{3} cos t) ) D. ( x=a(sqrt{3} sin t+cos t) ) | 11 |
| 398 | A particle moves in a circular path with a uniform speed. Its motion is: A. Periodic B. Oscillatory c. simple harmonic D. Angular simple harmonic | 11 |
| 399 | Which of the following quantities is non-zero at the mean position for a particle executing SHM? ( A cdot ) force B. acceleration c. velocity D. displacement | 11 |
| 400 | The maximum speed of a body vibrating under S.H.M. with time period of ( pi / 4 ) s amplitude ( 7 mathrm{cm} ) is ( A cdot 488 mathrm{cm} / mathrm{s} ) B. ( 56 mathrm{cm} / mathrm{s} ) c. ( 38.5 mathrm{cm} / mathrm{s} ) D. ( 5.5 mathrm{cm} / mathrm{s} ) | 11 |
| 401 | A particle starts from a point ( mathrm{P} ) at ( mathrm{a} ) distance of A/2 from the mean position 0 and travels towards left as shown in the figure. If the time period of ( mathrm{SHM} ) executed about 0 is ( mathrm{T} ) and amplitude ( mathrm{A} ) then the equation of the motion of particle is: This question has multiple correct options A ( cdot x=A sin left(frac{2 pi}{T} t+frac{pi}{6}right) ) B. ( x=A sin left(frac{2 pi}{T} t+frac{5 pi}{6}right) ) c. ( x=A cos left(frac{2 pi}{T} t+frac{pi}{6}right) ) D. ( x=A cos left(frac{2 pi}{T} t+frac{pi}{3}right) ) | 11 |
| 402 | A system of springs with their springs constants are as shown in figure. The frequency of oscillation of the mass ( m ) will be (assuming the springs to be massless A ( cdot frac{1}{2 pi} sqrt{frac{k_{1} k_{2}left(k_{3}+k_{4}right)}{left[left(k_{1}+k_{2}right)+left(k_{3}+k_{4}right)+k_{1} k_{4}right] m}} ) B ( cdot frac{I}{2 pi} sqrt{frac{k_{1} k_{2}left(k_{3}+k_{4}right)}{left[left(k_{1}+k_{2}right)+left(k_{3}+k_{4}right)+k_{1} k_{2}right] m}} ) c. ( frac{1}{2 pi} sqrt{frac{k_{1} k_{2}left(k_{3}+k_{4}right)}{left[left(k_{1}+k_{2}right)+left(k_{3}+k_{4}right)+k_{1} k_{2}right] m}} ) D ( frac{1}{2 pi} sqrt{frac{left(k_{1}+k_{2}right)+left(k_{3}+k_{4}right)+k_{1} k_{2}}{left[left(k_{1}+k_{2}right)+left(k_{3}+k_{4}right)+k_{1} k_{2}right] m}} ) | 11 |
| 403 | Q Type your question- connected to a fixed vertical wall and the other end has a block of mass ( mathrm{m} ) initially at rest on a smooth horizontal surface. The spring is initially in natural length. Now a horizontal force F acts on the block as shown. Then the maximum extension in spring is equal to the maximum compression in the spring. Reason To compress and to expand an ideal unstretched spring by equal amount, same work is to be done on the spring. | 11 |
| 404 | Fill in the blank A point source emits sound equally in all direction in a non-absorbing medium. Two point ( P ) and ( Q ) are at a distance of ( 9 m ) and ( 25 m ) respectively from the source. The ratio of the amplitudes of the waves at ( P ) and ( Q ) is.. | 11 |
| 405 | A particle is executing simple harmonic motion with an amplitude ( A ) and time period ( T . ) The displacement of the particles after ( 2 T ) period from its initial position is A . ( A ) в. ( 4 A ) c. ( 8 A ) D. zero | 11 |
| 406 | A stone tied at the end of string ( 80 mathrm{cm} ) long is whirled in a horizontal circle with a constant speed. If the stone makes 25 revolutions in 14 s, what is the magnitude of acceleration of the stone? A ( .90 m s^{-2} ) B. ( 100 mathrm{ms}^{-2} ) ( mathbf{c} cdot 110 m s^{-2} ) D. ( 120 mathrm{ms}^{-2} ) | 11 |
| 407 | Starting from an origin, a body oscillates simple harmonically with a period of ( 2 s . ) After what time will its kinetic energy be ( 75 % ) of the total energy? A . ( 1 / 6 s ) B. ( 1 / 12 s ) ( mathrm{c} cdot 1 / 3 s ) D. ( 1 / 4 s ) | 11 |
| 408 | Draw a sketch showing the displacement of a body executing damped vibrations against time. | 11 |
| 409 | ( boldsymbol{E} ) is kinetic energy of a simple harmonic oscillator at its mean position. The phase angle from mean position at which its kinetic energy is ( boldsymbol{E} / 2 ) is : A . ( pi / 5 ) rad в. ( pi / 4 ) rad c. ( pi / 3 ) rad D. None of the above | 11 |
| 410 | The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:- A. zero в. ( 0.5 pi ) ( c ) D. ( 0.707 pi ) | 11 |
| 411 | A particle performing linear simple harmonic motion has maximum velocity ( 25 mathrm{cm} / mathrm{s} ) and maximum acceleration of ( 100 mathrm{cm} / mathrm{s}^{2} ). Find the amplitude and period of oscillation. ( (boldsymbol{pi}=mathbf{3 . 1 4 2}) ) | 11 |
| 412 | A particle executes SHM whose instantaneous displacement is given by ( boldsymbol{x}=boldsymbol{A} cos (boldsymbol{omega} boldsymbol{t}+boldsymbol{pi} / 2) . ) What will be its acceleration at ( t=T / 4 ) ( mathbf{A} cdot a=0 ) B . ( a=2 omega^{2} A ) ( mathbf{c} cdot a=omega^{2} A ) D. ( a=omega^{2} A / 2 ) | 11 |
| 413 | A steamer moves with velocity ( 3 mathrm{km} / mathrm{h} ) in and against the direction of river water whose velocity is ( 2 mathrm{km} / mathrm{h} ). Calculate the total time for the total journey if the boat travels ( 2 mathrm{km} ) in the direction of a stream and then back to its place: A. 2 hrs B. 2.5 hrs c. 2.4 hrs D. 3 hrs | 11 |
| 414 | Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec) A ( cdot frac{pi}{2} ) B. ( 2 pi ) ( c ) D. ( frac{pi}{4} ) | 11 |
| 415 | A plot of ( f ) vs ( (1 / T) ) gives A. a straight line passing through the origin B. a straight line parallel to frequency axis C. a straight line perpendicular to frequency axis D. a straight line having an intercept along frequency axis | 11 |
| 416 | The unit ‘hertz’ is same as. A. second B. Second( ^{-1} ) c. Metre D. metre ( ^{-1} ) | 11 |
| 417 | The displacement of a particle undergoing ( S H M ) of time period ( T ) is given by ( boldsymbol{x}(boldsymbol{t})=boldsymbol{x}_{boldsymbol{m}} cos (boldsymbol{omega} boldsymbol{t}-boldsymbol{phi}) . ) The particle is at ( boldsymbol{x}=-boldsymbol{x}_{m} ) at time ( boldsymbol{t}=mathbf{0} ) The particle is at ( boldsymbol{x}=+boldsymbol{x}_{boldsymbol{m}} ) when ( mathbf{A} cdot t=0.25 T ) в. ( t=0.50 T ) c. ( t=0.75 T ) D. ( t=1.00 T ) | 11 |
| 418 | Apparent weight at bottom most line is A . 2 mg в. ( m g ) ( mathrm{c} .3 mathrm{mg} ) D. ( 4 mathrm{mg} ) | 11 |
| 419 | Two particle performed simple harmonic motion of same frequency an about same mean position. Their amplitude is same and is equal to ( boldsymbol{A} ) and their time period is ( T . ) If at ( t=0 ) separation is maximum and is ( A ), their separation at ( t=frac{T}{12} ) is:- A ( cdot quad A frac{sqrt{3}}{2} ) B. ( A ) c. ( frac{A}{sqrt{2}} ) D. ( frac{A}{2} ) | 11 |
| 420 | A body is executing simple harmonic motion of amplitude ( a ) and period ( T ) about the equilibrium position ( boldsymbol{x}=mathbf{0} ) Large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval ( boldsymbol{x} ) to ( boldsymbol{x}+|boldsymbol{d} boldsymbol{x}| ) is highest at A. ( x=pm a ) В. ( x=pm a / 2 ) c. ( x=0 ) D. ( x=pm a / sqrt{2} ) | 11 |
| 421 | A body executes simple harmonic motion. At a displacement ( x ), its potential energy is ( U_{1} . ) At a displacement ( y, ) its potential energy is ( U_{2} . ) What is the potential energy of the body at a displacement ( (x+y) ? ) A. ( U_{1}+U_{2} ) В ( cdot(sqrt{U_{1}}+sqrt{U_{2}})^{2} ) c. ( sqrt{U_{1}^{2}+U_{2}^{2}} ) D. ( sqrt{U_{1} U_{2}} ) | 11 |
| 422 | A particle executes simple harmonic motion between ( x=-A ) and ( x=+A ).The time taken by it to go from 0 to A /2 is ( T ) 1 and to go from ( A / 2 ) to ( A ) is ( T_{2} . ) Then ( A cdot T_{1}T_{2} ) ( mathrm{c} cdot mathrm{T}_{1}=mathrm{T}_{2} ) D. ( mathrm{T}_{1}=2 mathrm{T}_{2} ) | 11 |
| 423 | Frequency of oscillation of a body is 6 ( H z ) when force ( F_{1} ) is applied and ( 8 H z ) when ( F_{2} ) is applied. If both forces ( F_{1} & F_{2} ) are applied together then, the frequency of oscillation is : ( mathbf{A} cdot 14 H z ) в. ( 2 mathrm{Hz} ) c. ( 10 H z ) D. ( 10 sqrt{2} mathrm{Hz} ) | 11 |
| 424 | Fill in the blank A cylindrical resonance tube open at both ends fundamental frequency ( boldsymbol{F} ) in the air. Half-length of the tube is dipped vertically in the water. The fundamental frequency to the air column now is | 11 |
| 425 | Two springs of force constants ( mathbf{1 0 0 0} N / boldsymbol{m} ) and ( mathbf{2 0 0 0} boldsymbol{N} / boldsymbol{m} ) are stretched by same force. The ratio of their respective potential energies is A .2: 1 B. 1: 2 c. 4: 1 D. 1: 4 | 11 |
| 426 | A body undergoing ( S H M ) about the origin has its equation is given by ( boldsymbol{x}= ) ( 0.2 cos 5 pi t . ) Find its average speed in ( boldsymbol{m} / boldsymbol{s} ) from ( boldsymbol{t}=mathbf{0} ) to ( boldsymbol{t}=mathbf{0 . 7} boldsymbol{s e c} ) | 11 |
| 427 | Find force constant for small oscillation around equilibrium: A. ( 580 N m^{-1} ) B. ( 58 N m^{-1} ) c. ( 400 N m^{-1} ) D. none of these | 11 |
| 428 | Displacement versus time curve for a particle executing SHM is as shown in figure. At what points the velocity of the particle is zero? ( A cdot A, C, E ) B. В, D, F ( c cdot A, D, F ) D. ( mathrm{C}, mathrm{E}, mathrm{F} ) | 11 |
| 429 | When a particle oscillates simple harmonically, its kinetic energy varies periodically. If frequency of the particle is ( n, ) the frequency of the kinetic energy is: A ( cdot n / 2 ) в. ( n ) ( c cdot 2 n ) D. ( 4 n ) | 11 |
| 430 | An unhappy mouse of mass ( m_{0}, ) moving on the end of a spring of spring constant p is acted upon by a damping force ( F_{x}=-b v_{x} . ) For what value of ( b ) the motion is critically damped? A ( cdot b=sqrt{frac{p}{m_{0}}} ) B. ( b=2 sqrt{p m_{0}} ) ( ^{mathrm{c}}_{b}=sqrt{frac{p^{2}}{2 m_{0}}} ) D. ( b=sqrt{frac{p}{2 m_{0}}} ) | 11 |
| 431 | The amplitude of a particle executing SHM about 0 is ( 10 mathrm{cm} . ) Then: A. when the KE is 0.64 times of its maximum KE, its displacement is 6 cm from 0 B. its speed is half the maximum speed when its displacement is half the maximum displacement ( c . ) Both (a) and (b) are correct D. Both (a) and (b) are wrong | 11 |
| 432 | A particle of mass ( m ) is executing oscillations about the origin on the ( x- ) axis. It’s potential energy is ( boldsymbol{U}(boldsymbol{x})= ) ( boldsymbol{k}|boldsymbol{x}|^{3}, ) where ( boldsymbol{k} ) is a positive constant. If the amplitude of oscillation is ( a ), then its time period ( boldsymbol{T} ) is: A . proportional to ( frac{1}{sqrt{a}} ) B. independent of ( a ) c. proportional to ( sqrt{a} ) D. proportional to ( a^{3 / 2} ) | 11 |
| 433 | Displacement time equation of a particle executing SHM is ( boldsymbol{x}= ) ( 10 sin left(frac{pi}{3} t+frac{pi}{6}right) c m . ) The distance covered by particle in 3 seconds is A. ( 5 mathrm{cm} ) B. 20 cm c. ( 10 mathrm{cm} ) D. ( 15 mathrm{cm} ) | 11 |
| 434 | The maximum value attained by the tension in the string of a swinging pendulum is four times the minimum value it attains. There is no slack in the string. The angular amplitude of the pendulum is? A ( cdot 90^{circ} ) B. ( 60^{circ} ) ( c cdot 45^{circ} ) D. ( 30^{circ} ) | 11 |
| 435 | The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is ( 5 s . ) In another ( 10 s ) it will decrease to ( alpha ) times its original magnitude, where ( alpha ) equals. A . 0.81 B. 0.729 c. 0.6 D. 0.7 | 11 |
| 436 | Which of the following equation does not represent a SHM? ( mathbf{A} cdot cos omega t+sin omega t ) ( mathbf{B} cdot sin omega t-cos omega t ) c. ( 1-sin 2 omega t ) ( mathbf{D} cdot sin omega t+cos (omega t+alpha) ) | 11 |
| 437 | If a simple pendulum has significant amplitude (up to a factor of ( 1 / ) e of original) only in the period between ( t= ) ( 0 s ) to ( t=tau s, ) then ( tau ) may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with b’ as the constant proportionality, the average life time of the pendulum in seconds (assuming damping is small) in second :- A. ( frac{0.693}{b} ) B. ( b ) c. ( frac{1}{b} ) D. ( frac{2}{b} ) | 11 |
| 438 | Swinging of table fan is an example of which type of motion? A. Oscillatory motion B. One dimenetional motion c. Projectile motion D. None of the above | 11 |
| 439 | The displacement of an oscillating particle varies with time ‘t’ (in seconds) ( operatorname{according} operatorname{as} x=2 cos (0.5 pi t) ) metre Then amplitude and time period are : ( A cdot 1 m, 4 s ) B. 2 ( m, 4 ) s ( c cdot 2 m, 2 s ) D. ( 1 mathrm{m}, 2 mathrm{s} ) | 11 |
| 440 | The displacement of a body executing SHM is given by ( x=A sin left(2 pi t+frac{pi}{3}right) . ) The first time from ( t=0 ) when the velocity is maximum is: A. 0.33 sec B. 0.16 sec c. ( 0.25 mathrm{sec} ) D. ( 0.5 mathrm{sec} ) | 11 |
| 441 | In damped vibrations, as time progresses, amplitude of oscillation A. decreases B. increases c. Remains same D. Data insufficient | 11 |
| 442 | A particle executes ( S . H . M . ) of amplitude ( a ) 1- At what distance from mean position is its kinetic energy equal to its potential energy 2- At what points is its speed half the maximum speed | 11 |
| 443 | Displacement of a particle executing simple harmonic motion is represented by ( boldsymbol{Y}=mathbf{0 . 0 8} sin left(3 pi t+frac{pi}{4}right) ) metre. Then calculate:- (a) Time period. (b) Initial phase (c) Displacement from mean position at ( t=frac{7}{36} ) sec. | 11 |
| 444 | Classify the following as linear, circular, vibratory or oscillatory motion. ¡) The motion of a swing. ii) The motion of earth around the sun iii) The motion of a cyclist on a plain road. iv) The motion of a falling stone v) The motion of a plucked string of a sitar. | 11 |
| 445 | A particle executes SHM in a straight line. The maximum speed of the particle during its motion is ( V_{text {max}} ). Then the average speed of the particle during the SHM is : A ( cdot frac{V_{m}}{pi} ) в. ( frac{V_{m}}{2 pi} ) c. ( frac{2 V_{m}}{pi} ) D. ( frac{3 V_{m}}{pi} ) | 11 |
| 446 | Assertion:In simple harmonic motion A is the amplitude of oscillation. If ( t_{1} ) be the time to reach the particle from mean position to ( frac{A}{sqrt{2}} ) and ( t_{2} ) the time to reach from ( frac{boldsymbol{A}}{sqrt{mathbf{2}}} ) to ( mathbf{A}, ) then ( boldsymbol{t}_{mathbf{1}}=frac{boldsymbol{t}_{mathbf{2}}}{sqrt{mathbf{2}}} ) Reason : Equation of motion for the particle starting from mean position is ( operatorname{given} operatorname{by} x=A cos (omega t) ) and of the particle starting from extreme position is given by ( boldsymbol{A} sin (boldsymbol{omega} boldsymbol{t}) ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 11 |
| 447 | The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of ( 1.0 mathrm{m} . ) If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what it its maximum speed? | 11 |
| 448 | A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy(K.E.) and total energy (T.E.) are measured as function of displacement x. Which of the following statement is true? A. K.E. is maximum when ( x=0 ) B. T.E. is zero when ( x=0 ) c. K.E. is maximum when x is maximum D. P.E. is maximum when ( x=0 ) | 11 |
| 449 | Which of the following regarding oscillatory motion is true? A. Motion of the earth is periodic but not oscillatory because it is not to and fro. B. Quivering of the string of the musical instrument is an example of oscillatory motion C. Motion of the earth is periodic and oscillatory motion because it is not to and fro. D. None of the above | 11 |
| 450 | A particle executes S.H.M. of period ( 6.28 s ) and amplitude ( 3 mathrm{cm} . ) Its maximum acceleration is A ( .2 .14 mathrm{cm} / mathrm{s}^{2} ) в. 3 ст / ( s^{2} ) c. ( 9.42 mathrm{cm} / mathrm{s}^{2} ) D. ( 1.5 mathrm{cm} / mathrm{s}^{2} ) | 11 |
| 451 | Which of the following results in more energy in a wave? A. a smaller wavelength B. a lower frequency c. a shallower amplitude D. a lower speed | 11 |
| 452 | If the displacement equation of a particle from its mean position is given as ( boldsymbol{y}=mathbf{0 . 2} sin (mathbf{1 0} boldsymbol{pi} boldsymbol{t}+ ) ( 1.5 pi) cos (10 pi t+1.5 pi) ) then, the motion of particle is A. Non-periodic B. Periodic but not SHM c. SHM with period 0.2 D. SHM with period 0.1 s | 11 |
| 453 | A spring mass system oscillates with a time period ( T_{1} ) when a certain mass is attached to the spring. When a different mass is attached after removing the first mass the time period becomes ( T_{2} ) When both the masses are attached to the free end of the spring with the other end fixed, the time period of oscillation is : ( mathbf{A} cdot sqrt{T_{1} T_{2}} ) в. ( T_{1}+T_{2} ) c. ( frac{T_{1}+T_{2}}{2} ) D. ( sqrt{T_{1}^{2}+T_{2}^{2}} ) | 11 |
| 454 | In a particle executing SHM, the following parameter at equilibrium will have its A. Kinetic energy as maximum B. Potential energy as maximum C. Kinetic energy as minimum D. None of the above | 11 |
| 455 | Consider the situation shown in figure. Show that if the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. Calculate the time period. uculate | 11 |
| 456 | ( i= ) ( k ) ( L ) | 11 |
| 457 | Choose the correct option which describe the simple harmonic motion. I. The acceleration is constant. II. The restoring force is proportional to the displacement III. The frequency is independent of the amplitude. A. Il only B. I and II only c. I and III only D. Il and III only E. I, II and III | 11 |
| 458 | In a simple harmonic motion ( (mathrm{SHM}) ) which of the following does not hold? A. The force on the particle is maximum at the ends B. The acceleration is minimum at the mean postion C. The potential energy is maximum at the mean position D. The kinetic energy is maximum at the mean position | 11 |
| 459 | rooe vaurauestion a, which of the following graph is | 11 |
| 460 | In a SHM, when the displacement is one half the amplitude, what fraction of the total energy is kinetic? A. zero B. ( 1 / 4 ) ( c cdot 1 / 2 ) D. 3/4 | 11 |
| 461 | Equation of SHM is ( x=10 sin 10 pi t . ) Then the time period is, ( x ) is in ( mathrm{cm} ) and ( mathrm{t} ) is in ( sec ) ( A cdot 10 pi ) B. 0.2 sec c. ( 0.1 mathrm{sec} ) D. 2 sec | 11 |
| 462 | The phase difference between the two simple harmonic oscillations, ( boldsymbol{y}_{1}= ) ( frac{1}{2} sin omega t+left(frac{sqrt{3}}{2}right) cos omega t ) and ( y_{2}= ) ( sin omega t+cos omega t ) is A ( cdot frac{pi}{6} ) B. ( -frac{pi}{6} ) c. ( frac{pi}{12} ) D. ( frac{7 pi}{12} ) | 11 |
| 463 | The torque equation for a physical pendulum making an angle ( theta ) with the vertical is given by A ( cdot d^{2} theta / d t^{2}=-(m g L / I) cos theta ) B . ( d^{2} theta / d t^{2}=-(m g L / I) sin theta ) C ( cdot d^{2} theta / d t^{2}=-(m g L / 2 I) sin theta ) D ( cdot d^{2} theta / d t^{2}=-(m g L / 2 I) cos theta ) | 11 |
| 464 | Why does the amplitude of a vibrating body continuously decrease during damped vibrations? | 11 |
| 465 | The periodic time of a particle doing simple harmonic motion is 4 s. The time taken for it to go from its mean position to half the maximum displacement (amplitude) is. ( mathbf{A} cdot 2 s ) B . ( 1 s ) c. ( frac{2}{3} s ) D. ( frac{1}{3} s ) | 11 |
| 466 | Which of the following statement(s) is/are correct regarding the amplitude vs frequency curve for a driven system? This question has multiple correct options A. Smaller the damping, taller and narrower the resonance peak B. The amplitude tends to infinity when it equals to natural frequency results in zero damping C. The resonant amplitude decreases with increasing damping D. Small damping results in driving frequency almost equal to natural frequency | 11 |
| 467 | When a body is in SHM, match the statements in Column A with that in Column B Column A Column B a) Velocity is maximum e) At half of the amplitude b) Kinetic energy is At the mean position ( 3 / 4 ) th of total energy c) Potential energy is g) At extreme position ( 3 / 4 ) th of total energy d) Acceleration is h) ( mathrm{At} sqrt{3} / 2 ) times amplitude maximum A. a-f, b-e, c-h, d-g B. a-e, b-f, c-g, d-h c. a-g, b-h, c-e, d-f D. a-h, b-e, c-f, d-e | 11 |
| 468 | A leapord leaps a distance of 1 metre in a straight line, 20 times in 1 minute, while catching a prey, then A. Time period and frequency cannot be determined since the leopard is not returning to its original position B. The product of its time period and frequency will be unity c. The time period is infinity D. The frequency is zero | 11 |
| 469 | How many times will light travel around the Earth in one second? A. 7 times B. 5 times c. 3 times D. 9 times | 11 |
| 470 | Which of the following quantities are always zero in SHM? This question has multiple correct options A ( cdot vec{F} times vec{a} ) В. ( vec{v} times vec{r} ) c ( . vec{a} times vec{r} ) D. ( vec{F} times vec{r} ) | 11 |
| 471 | Two particles ( P ) and ( Q ) describe simple harmonic motions of same period, same amplitude, along thesame line about the same equilibrium position 0 When ( P ) and ( Q ) are on opposite sides of 0 at the samedistance from 0 they have the same speed of ( 1.2 mathrm{m} / mathrm{s} ) in the same direction, when their displacements arethe same they have the same speed of ( 1.6 mathrm{m} / mathrm{s} ) in opposite directions. The maximum velocity in ( mathrm{m} / mathrm{s} ) ofeither particle is A . 2.8 B. 2.5 ( c cdot 2.4 ) D. 2 | 11 |
| 472 | one end has a small block of mass ( m ) and charge ( q ) is attached at the other end. The block rests over a smooth horizontal surface. A uniform and constant magnetic field ( B ) exists normal to that plane of paper as shown in figure. An electric field ( overrightarrow{boldsymbol{E}}=boldsymbol{E}_{0} hat{boldsymbol{i}}left(boldsymbol{E}_{mathbf{0}}right. ) is a positive constant) is switched on at ( boldsymbol{t}=mathbf{0} ) sec. The block moves on horizontal surface without ever lifting off the surface. Then the normal reaction acting on the block is A. Maximum at extreme position and minimum at mean position B. Maximum at mean position and minimum at extreme position C. is uniform throughout the motion D. is both maximum and minimum at mean position | 11 |
| 473 | A Simple harmonic oscillator has a period of ( 0.01 s ) and an amplitude of ( 0.2 m . ) The magnitude of the velocity in ( m ) ( sec ^{-1} ) at the mean position will be A ( .20 pi ) в. 100 c. ( 40 pi ) D. ( 100 pi ) | 11 |
| 474 | Variations of acceleration a of a particle executing SHM with displacement x is: | 11 |
| 475 | The energy of a particle executing simple harmonic motion is given by the equation ( boldsymbol{E}=boldsymbol{A} boldsymbol{x}^{2}+boldsymbol{B} boldsymbol{v}^{2} ) where ( boldsymbol{x} ) is the displacement from mean position ( boldsymbol{x}= ) 0 and ( v ) is the velocity of the particle at ( x . ) Find the amplitude of ( S . H . M ) | 11 |
| 476 | Which of the following is not simple harmonic function? A ( cdot y=a sin 2 omega t+b cos ^{2} omega t ) B. ( y=a sin omega t+b cos 2 omega t ) c. ( y=1-2 sin ^{2} omega t ) D ( cdot y=(sqrt{a^{2}+b^{2}}) sin omega t cos omega t ) | 11 |
| 477 | A particle performing SHM has time period ( frac{2 pi}{sqrt{3}} ) and path length ( 4 mathrm{cm} . ) The displacement from mean position at which acceleration is equal to velocity is ( A cdot 0 c m ) B. ( 0.5 mathrm{cm} ) ( c cdot 1 c m ) D. ( 1.5 mathrm{cm} ) | 11 |
| 478 | An object of mass ( 0.2 mathrm{kg} ) executes SHM along the ( x ) -axis with frequency ( (25 / pi) H z . A t ) the point ( x=0.04 m ) the object has ( mathrm{KE} 0.5 mathrm{J} ) and PE 0.4 J.The amplitude of oscillation (in cm) is | 11 |
| 479 | If ( x, v ) and ( a ) denote the displacement the velocity and the acceleration of a particle executing simple harmonic motion of time period ( T, ) then, which of the following does not change with time? в. ( frac{mathrm{aT}}{mathrm{x}} ) c. ( a T+2 pi v ) D. ( frac{mathrm{aT}}{mathrm{v}} ) | 11 |
| 480 | For a particle in SHM, if the amplitude of the displacement is ( alpha ) and the amplitude of velocity is ( v, ) the amplitude of acceleration is A ( . v alpha ) в. ( frac{v^{2}}{alpha} ) c. ( frac{v^{2}}{2 alpha} ) D. ( underline{v} ) | 11 |
| 481 | The kinetic energy of a simple pendulum may be decreased by: A. increasing the mass of the bob B. increasing the thickness of the string C. decreasing the length of the string D. increasing the length of the string E. decreasing the displacement of the string | 11 |
| 482 | The frequency ( f ) of vibration of mass ( m ) suspended from a spring of spring contact ( k ) is given by ( boldsymbol{f}=boldsymbol{c} boldsymbol{m}^{boldsymbol{x}} boldsymbol{k}^{boldsymbol{y}} ) Where ( c= ) dimensionless constant then find the values of ( x ) and ( y ) | 11 |
| 483 | The work done in stretching a spring of force constant ( mathrm{k} ) from ( boldsymbol{y}_{1} ) and ( boldsymbol{y}_{2} ) from mean position will be A ( cdot frac{k}{2}left(y_{2}^{2}-y_{1}^{2}right) ) B. ( frac{k}{2}left(y_{1}^{2}-y_{2}^{2}right) ) ( mathbf{c} cdot kleft(y_{2}-y_{1}right) ) D. zero | 11 |
| 484 | The correct relation between frequency and time period is ( ^{mathbf{A}} cdot T=frac{1}{f} ) B ( cdot T=sqrt{frac{1}{f}} ) c. ( T=frac{1}{2 f} ) D. ( T=frac{2}{f} ) | 11 |
| 485 | Two simple Harmonic Motions of angular frequency 100 and 1000 rad ( S^{-1} ) have the same displacement amplitude. The ratio of their maximum accelerations is : A ( cdot 1: 10^{3} ) B ( cdot 1: 10^{4} ) c. 1: 10 D. ( 1: 10^{2} ) | 11 |
| 486 | ( frac{1}{0} ) ( frac{partial}{t} ) | 11 |
| 487 | If the displacement, velocity and acceleration of particle in ( S H M ) are ( mathbf{1} boldsymbol{c m}, mathbf{1} boldsymbol{c m} / boldsymbol{s e c} ) and ( mathbf{1} boldsymbol{c m} / boldsymbol{s e c}^{mathbf{2}} ) respectively its time period (in secs) will be: A . ( pi ) B. ( 0.5 pi ) ( c .2 pi ) D. ( 1.5 pi ) | 11 |
| 488 | A particle of ass ( mathrm{m} ) and charge ( mathrm{Q} ) is placed in an electric field E which varies with time ( t ) ass ( E=E_{0} ) sin ( omega t . ) It will undergo simple harmonic motion of amplitude A ( cdot frac{Q E_{0}^{2}}{m omega^{2}} ) в. ( frac{Q E_{0}}{m omega^{2}} ) c. ( sqrt{frac{Q E_{0}}{m omega_{2}}} ) D. ( frac{Q E_{0}}{m omega} ) | 11 |
| 489 | The potential energy of a simple harmonic oscillator of mass ( 2 mathrm{kg} ) in its mean position is 5 J. If its totalenergy is ( 9 mathrm{J} ) and its amplitude is ( 0.01 mathrm{m}, ) its time period would be A. ( pi / 10 ) sec B . ( pi / 20 mathrm{sec} ) c. ( pi / 50 ) sec D. ( pi / 100 ) sec | 11 |
| 490 | Match the items in List 1 with the items in List 2 | 11 |
| 491 | A student says that he had applied a force ( F=-k sqrt{x} ) on a particle and the particle moves in simple harmonic motion. He refuses to tell whether ( k ) is a constant or not. Assume that he has worked only with positive ( x ) and no other force acted on the particle. A . As ( x ) increases ( k ) increases B. As ( x ) increases ( k ) decreases C. As ( x ) increases ( k ) remains constant D. The motion cannot be simple harmonic | 11 |
| 492 | Earth revolves around sun in 365 days.Calculate its angular speed. | 11 |
| 493 | A rope, under a tension of ( 200 N ) and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by ( : boldsymbol{y}=mathbf{0 . 1} sin left(frac{boldsymbol{pi} boldsymbol{x}}{mathbf{2}}right) sin 12 boldsymbol{pi} boldsymbol{t} ) Where ( x=0 ) at one end of the rope, ( x ) is in meters and ( t ) is in seconds. The speed of the progressive waves ( (operatorname{in} m / s) ) on the rope is | 11 |
| 494 | Two particles execute SHM of same amplitude and same time period, about same mean position but with a phase difference between them. At an instant they are found to cross each other at ( x=+frac{A}{3} . ) The phase difference between them is: ( mathbf{A} cdot 2 cos ^{-1}left[frac{1}{5}right] ) B ( cdot 2 sin ^{-1}left[frac{1}{5}right] ) C ( cdot 2 cos ^{-1}left[frac{1}{3}right] ) D. ( 2 sin ^{-1}left[frac{1}{3}right] ) | 11 |
| 495 | The particle is executing S.H.M. on a line ( 4 mathrm{cms} ) long. If its velocity at its mean position is ( 12 mathrm{cm} / mathrm{sec}, ) its frequency in Hertz will be : A ( cdot frac{2 pi}{3} ) в. ( frac{3}{2 pi} ) c. D. ( frac{3}{pi} ) | 11 |
| 496 | A simple harmonic oscillator has a period ( T ) and energy E. The amplitude of the oscillator is doubles. Choose the correct answer. A. Period and energy get doubled B. Period gets doubled while energy remains the same. c. Energy gets doubled while period remains the same D. Period remains the same and energy becomes four times | 11 |
| 497 | The period of the SHM of a particle with the maximum velocity ( 50 mathrm{cm} / mathrm{s} ) and maximum acceleration ( 10 mathrm{cm} / mathrm{s}^{2} ) is A . ( 31.42 s ) B. ( 6.284 s ) c. ( 3.142 s ) D. ( 0.3124 s ) | 11 |
| 498 | Calculate the velocity of the bob of a simple pendulum at its mean position if it is able to rise to a vertical height of ( 10 mathrm{cm} . ) Given: ( g=980 mathrm{cms}^{-2} ) | 11 |
| 499 | A particle executes simple harmonic motion along a straight line with mean position at ( x=0 ) and period of ( 20 s ) and amplitude of ( 5 mathrm{cm} . ) The shortest time taken by the particle to go from ( boldsymbol{x}=mathbf{4 c m} ) to ( boldsymbol{x}=-mathbf{3 c m} ) is A . ( 4 s ) B. 7 s ( c .5 s ) D. ( 6 s ) | 11 |
| 500 | Statement 1: In simple harmonic motion, the motion is to and fro and periodic. Statement 2: Velocity of the particle | 11 |
| 501 | The simple harmonic motion of a particle is given by ( boldsymbol{x}=boldsymbol{a} sin 2 boldsymbol{pi} boldsymbol{t} . ) Then the location of the particle from its mean position at a time ( 1 / 8 ) th of a second is: ( mathbf{A} cdot boldsymbol{a} ) B. ( frac{a}{2} ) c. ( frac{a}{sqrt{2}} ) D. ( frac{a}{4} ) E ( cdot frac{a}{8} ) | 11 |
| 502 | A tunnel is dug along the diameter of the earth. A mass ( m ) is dropped into it. How much time does it take to cross the earth? A. 169.2 minutes B. 84.6 minutes c. 21.2 minutes D. 42.3 minutes | 11 |
| 503 | If the displacement of a particle executing SHM is given by ( boldsymbol{y}= ) ( 0.30 sin (220 t+0.64) ) in metre, then the frequency and maximum velocity of the particle are A ( .35 H z, 66 m / s ) B. ( 45 H z, 66 m / s ) ( mathbf{c} .58 H z, 113 m / s ) D. ( 35 H z, 132 m / s ) | 11 |
| 504 | Two particles are in SHM along same line. Time period of each is ( T ) and amplitude is ( A . ) After how much time will they collide if at time ( t=0 . ) First particle is at ( x_{1}=+frac{A}{2} ) and moving towards positive ( x- ) axis and second particle is at ( x_{2}=-frac{A}{sqrt{2}} ) and moving towards negative ( x- ) axis. A ( cdot frac{18}{19} T ) в. ( frac{19}{48} mathrm{T} ) c. ( frac{48}{19} T ) D. ( frac{19}{18} mathrm{T} ) | 11 |
| 505 | Restoring force in the SHM is A. conservative B. nonconservative C . frictional D. centripetal | 11 |
| 506 | The displacement of a particle is represented by the equation ( y= ) ( 3 cos left[frac{pi}{4}-2 omega tright] ) The motion is: A. non- periodic B. periodic but not simple harmonic c. simple harmonic with period 2 piw D. simple harmonic with period ( left(frac{pi}{omega}right) ) | 11 |
| 507 | In a Vander waals interaction: ( boldsymbol{U}= ) ( boldsymbol{U}_{0}left[left(frac{boldsymbol{R}_{0}}{boldsymbol{r}}right)^{12}-boldsymbol{2}left(frac{boldsymbol{R}_{0}}{boldsymbol{r}}right)^{6}right] ) A small displacement x is given from equilibrium position ( r=R_{0} . ) Find the approximate PE function. A ( cdot frac{36 U_{0}}{R_{0}^{2}} x^{2}-U_{0} ) в. ( frac{24 U_{0}}{R_{0}} x-U_{0} ) c. ( frac{96 U_{0}}{R_{0}^{2}}-U_{0} ) D. none of these | 11 |
| 508 | A particle executes SHM on a straight line path. The amplitude of oscillation is ( 2 mathrm{cm} . ) When the displacement of the particle from the mean position is ( 1 mathrm{cm} ) the numerical value of the magnitude of acceleration is equal to the numerical value of the magnitude of the velocity. The frequency of SHM is: ( A cdot 2 pi sqrt{3} ) B. ( frac{2 pi}{sqrt{3}} ) ( c cdot frac{sqrt{3}}{2 pi} ) D. ( frac{1}{2 pi sqrt{3}} ) | 11 |
| 509 | In forced vibration ( boldsymbol{m}=mathbf{1 0 g m}, boldsymbol{f}= ) ( 100 H z ) and driver force ( F= ) ( 100 cos (20 pi t) ) then what amplitude of particle. | 11 |
| 510 | A mass ( m, ) which is attached to a spring with constant ( k, ) oscillates on a horizontal table, with amplitude ( A . ) At an instant when the spring is stretched by ( sqrt{3} A / 2, ) a second mass ( m ) is dropped vertically onto the original mass and immediately sticks to it. What is the amplitude of the resulting motion? A ( cdot frac{sqrt{3}}{2} A ) B. ( sqrt{frac{7}{8}} A ) c. ( sqrt{frac{13}{16}} A ) D. ( sqrt{frac{2}{3}} A ) | 11 |
| 511 | If a simple pendulum of length L has maximum angular displacement ( boldsymbol{theta} ) then the maximum kinetic energy of the bob of mass ( mathrm{m} ) is A ( cdot frac{1}{2} m(L / g) ) В. ( m g L(1-cos theta) ) c. ( (m g L sin theta) / 2 ) D. ( m g / 2 L ) | 11 |
| 512 | Three similar oscillators, ( A, B, C ) have the same small damping constant ( r ) but different natural frequencies ( omega_{0}= ) ( (k / m)^{frac{1}{2}}: 1200 H z, 1800 H z, 2400 H z .1 f ) all three are driven by the same source at ( 1800 H z, ) which statement is correct for the phases of the velocities of the three? A ( cdot phi_{A}=phi_{B}=phi_{c} ) в. ( phi_{A}<phi_{B}=0phi_{B}=0>phi_{c} ) D ( cdot phi_{A}>phi_{B}>0>phi_{c} ) | 11 |
| 513 | A weakly damped harmonic oscillator is executing resonant oscillations. The phase difference between the oscillator and the external periodic force is: A. zero B . ( pi / 4 ) c. ( pi / 2 ) ( D ) | 11 |
| 514 | Which of the following quantity does not change due to damping of oscillations? A. Angular frequency B. Time period c. Initial phase D. Amplitude | 11 |
| 515 | Choose the correct statement with respect to the graph given(consider the case of simple mass spring, ( ) ) A. A force must be acting on the system B. B represent kinetic energy C. A represent potential energy D. none of the above | 11 |
| 516 | A light spiral spring supports a 200 g weight at its lower end. It oscillates up and down with a period of 1 sec. How much weight (gram) must be removed from the lower end to reduce the period to 0.5 sec.? A . 200 B. 50 ( c .53 ) D. 100 | 11 |
| 517 | The damping force on a oscillator is directly proportional to the velocity. The Unit of the constant of proportionality are: A. ( mathrm{kgs}^{-1} ) B. kgs c. ( operatorname{kgms}^{-2} ) D. kgms | 11 |
| 518 | The equation of a wave is given by ( boldsymbol{Y}= ) ( A sin omegaleft(frac{x}{v}-kright), ) where ( omega ) is the angular velocity and ( v ) is the linear velocity. Find the dimension of ( k ? ) | 11 |
| 519 | The shortest distance travelled by a particle performing SHM from its mean position in ( operatorname{secs} text { is } 1 / sqrt{(} 2) ) of its amplitude. Find its time period: ( A cdot 8 ) B. 16 ( c cdot 4 ) D. | 11 |
| 520 | Vibrations, whose amplitudes of oscillation decrease with time, are called: A. free vibrations B. forced vibrations c. damped vibrations D. sweet vibrations | 11 |
| 521 | If a body moves back and forth repeatedly about a mean position, it is said to possess A. rotatory motion B. projectile motion c. oscillatory motion D. Reciprocating motion | 11 |
| 522 | A particle of mass is executing oscillations about the origin on the ( x ) axis. Its potential energy is ( V(x)= ) ( k|x|^{3}, ) where ( k ) is a positive constant. If the amplitude of oscillation is ( a ), then its time period ( T ) is proportional A . proportional to ( frac{1}{sqrt{a}} ) B. proportional to ( sqrt{a} ) C . Independent ( a^{frac{3}{2}} ) D. None of these | 11 |
| 523 | A particle in linear SHM performs ( 30 s c / s e c . ) Its velocity is ( 0.120 m / s ) when it passes through the middle of its path. The length of path is A. ( 0.012 mathrm{cm} ) B. ( 3.2 mathrm{cm} ) c. ( 0.04 c m ) D. ( 1.2 mathrm{cm} ) | 11 |
| 524 | The equation of motion of a particle executing SHM is ( left(frac{d^{2} x}{d t^{2}}right)+k x=0 . ) The time period of the particle will be ? A ( cdot frac{2 pi}{sqrt{k} pi} ) в. ( 2 pi ) ( c cdot 2 pi k ) D. ( 2 pi sqrt{k} ) | 11 |
| 525 | The ratio of ( x_{1} / x_{2} ) is ( A cdot 2 ) B. ( frac{1}{2} ) ( c cdot sqrt{2} ) D. ( frac{1}{sqrt{2}} ) | 11 |
| 526 | The vibrations of a body which take place under the influence of an external periodic force acting on it are called A. Forced vibrations B. Free vibration c. Damped vibrations D. All | 11 |
| 527 | A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of ( 10^{12} / ) sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver ( =108 ) and Avagadro number ( =mathbf{6 . 0 2} times mathbf{1 0}^{mathbf{2 3}} mathbf{g} mathbf{m} mathbf{m o l}^{-mathbf{1}} ) ). A. ( 2.2 mathrm{N} / mathrm{m} ) в. ( 5.5 mathrm{N} / mathrm{m} ) c. ( 6.4 mathrm{N} . mathrm{m} ) D. ( 7.1 mathrm{N} / mathrm{m} ) | 11 |
| 528 | A particle executing SHM is described by the equation ( boldsymbol{x}=boldsymbol{A} boldsymbol{e}^{boldsymbol{omega} t} ). Will this particle describe SHM. A. It dosen’t execute SHM B. It executes SHM only for the first time c. It executes SHM from ( x=0 ) to ( x=A ) D. It executes SHM from ( x=0 ) to ( x=-A ) | 11 |
| 529 | A particle of mass ( 2 mathrm{kg} ) is moving on a straight line under the action of force ( mathrm{F}= ) ( (8-2 x) ) newton along the ( x ) -axis where ( x ) is the coordinates on ( x ) -axis If a particle is released from ( x=7 ) m then for the subsequent motion match the following (all values in Column II are in S.I units) | 11 |
| 530 | For the given experimental setup, the oscillation frequency of the mass is ( boldsymbol{f} cdot boldsymbol{S}_{1} ) and ( boldsymbol{S}_{2} ) are identical springs. One spring is now removed, then frequency will become. A . ( 2 f ) B. ( frac{f}{sqrt{2}} ) c. ( frac{f}{sqrt{3}} ) D. ( sqrt{2} f ) | 11 |
| 531 | A spring mass system is in simple harmonic motion. Immediately after the upward extreme end point, which of the following increase? A. Kinetic energy, gravitational potential energy, elastic potential energy B. Kinetic energy and gravitational potential energy c. Kinetic energy only D. Gravitational potential energy and elastic potential energy E. Elastic potential energy only | 11 |
| 532 | A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude A.lts speed as it passes through the equilibrium position is V.If extended 2A and released, the speed of the mass passing through the equilibrium position will be A ( .2 v ) B. 4V c. v/2 D. V/4 | 11 |
| 533 | Which of the following motions is not simple harmonic? A. Vertical oscillations of a spring B. Motion of a simple pendulum c. Motion of planet around the sun D. oscillation of liquid in a U-tube | 11 |
| 534 | Find the natural frequency of oscillation of the system as shown in figure. Pulleys are massless and frictionless. Spring and string are also massless. A ( cdot frac{pi}{2} ) B. ( sqrt{pi} ) ( c cdot sqrt{frac{1}{pi}} ) ( D ) | 11 |
| 535 | Obtain an expression for potential energy of a particle performing simple harmonic motion. Hence evaluate the potential energy (a) at mean position and (b) at extreme position. A horizontal disc is freely rotating about a transverse axis passing through its centre at the rate of 100 revolutions per minute. A 20 gram blob of wax falls on the disc and sticks to the disc at a distance of ( 5 mathrm{cm} ) from its axis. Moment of inertia of the disc about its axis passing through its centre of mass is ( 2 times 10^{-4} k g m^{2} . ) Calculate the new frequency of rotation of the disc. | 11 |
| 536 | A block is attached to an unstretched vertical spring and released from rest. As a result of This the block comes down due to its weight. stops momentarily, and then bounces back. Finally, the block starts oscillating up and down. M Fig. 6.389 During oscillations, match Column I with Column II: Column I Column II When the block is at its a. Acceleration is in maximum downward upward direction. displacement position (may be known as extreme position) When the block is at its b. Acceleration is in equilibrium position downward direction. When the block is somewhere c. Acceleration is zero. between equilibrium position and downward extreme position When the block is above d. Velocity may be equilibrium position but in upward or in below the initial unstretched downward direction. position . ! iv. | 11 |
| 537 | The time period of a particle in SHM is 12 sec. At ( t=0, ) it is at mean position. The ratio of the distance travelled in the second and sixth second is. A ( cdot(sqrt{3-1}) ) B. ( (sqrt{2-1}) ) c. ( frac{1}{(sqrt{2}-1)} ) D. ( frac{1}{sqrt{2}} ) | 11 |
| 538 | The periodic vibrations of a body of decreasing amplitude in the presence of resistive force on it are called A. Forced vibrations B. Free vibration c. Damped vibrations D. All | 11 |
| 539 | toppr ०६ Q Type your question spring is unsurtecritu. Next with the blue mass hanging, the spring stretches. Finally, with the red mass hanging, the | 11 |
| 540 | A particle doing S.H.M. having amplitude ( 5 mathrm{cm}, ) mass ( 5 mathrm{kg} ) and angular frequency 5 is 1 cm from mean position. Find potential energy and kinetic energy. ( mathbf{A} cdot K cdot E .=150 times 10^{-4} J . P . E .=6.25 times 10^{-4} J ) B . ( K . E .=6.25 times 10^{-4} J . P . E .=150 times 10^{-3} J ) C ( . K . E .=6.25 times 10^{-4} J . P . E .=6.25 times 10^{-4} J ) D . ( K . E .=150 times 10^{-3} mathrm{J.P.E.}=150 times 10^{-4} mathrm{J} ) | 11 |
| 541 | What is the number of degrees of freedom of an oscillating simple pendulum? A. more than three B. 3 ( c cdot 2 ) D. | 11 |
| 542 | A particle is acted simultaneously by mutually perpendicular simple harmonic motions ( x=a c o s omega t ) and ( y= ) asinwt. The trajectory of motion of the particle will be A. an ellipse B. a parabola c. a circle D. a straight line | 11 |
| 543 | Assertion Motion of a ball bouncing elastically in vertical direction on a smooth horizontal floor is a periodic motion, but not an SHM. Reason Motion is SHM when restoring force is proportional to displacement from mean position. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 11 |
| 544 | An ant moves along a circle 5 times in 1 minute with a uniform speed and a lizard moves in a straight line to and fro 5 times in 5 minutes. Then A. The motion of the ant is uniformly accelerated but non periodic B. The motion of the lizard is uniformly accelerated but non periodic C. Both ant and lizard are periodic D. Both ant and lizard are accelerated and their direction is along the direction of velocity | 11 |
| 545 | The displacement of a particle varies with time according to the relation ( y= ) asinomegat ( + )bcosomegat. A. The motion is oscillatory but not SHM. B. The motion is SHM with amplitude a + b. c. The motion is SHM with amplitude ( a^{2}+b^{2} ) D. The motion is SHM with amplitude ( sqrt{a^{2}+b^{2}} ) | 11 |
| 546 | A force acts on a ( 30 mathrm{gm} ) particle in such a way that the position of the particle as a function of time is given by ( boldsymbol{x}=mathbf{3} boldsymbol{t}- ) ( 4 t^{2}+t^{3}, ) where ( x ) is in metres and ( t ) is in seconds. The work done during the first 4 second is? A .5 .28 B. 450mJ c. ( 490 mathrm{mJ} ) D. 530 mJ | 11 |
| 547 | A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude? A ( cdot(1 / 2) s ) в. ( (1 / 3) s ) c. ( (1 / 4) s ) D. ( (1 / 6) s ) | 11 |
| 548 | Instantaneous displacement of a particle is ( y=4 cos ^{2}left(frac{t}{2}right) sin 1000 t ) The particle is displaced due to: A. only one wave B. two waves c. three waves D. four waves | 11 |
| 549 | Displacement time equation of a particle SHM is ( boldsymbol{x}= ) ( mathbf{1 0} sin left(frac{boldsymbol{pi}}{mathbf{3}} boldsymbol{t}+frac{boldsymbol{pi}}{mathbf{6}}right) boldsymbol{c m} . ) The distance covered by particle in ( 3 s ) is A . ( 5 mathrm{cm} ) B. ( 20 mathrm{cm} ) c. ( 10 mathrm{cm} ) D. ( 15 mathrm{cm} ) | 11 |
| 550 | A stone is swinging in a horizontal circle of diameter ( 0.8 ~ m ) at 30 rev / min. A distant light causes a shadow of the stone on a nearly wall. The amplitude and period of the SHM for the shadow of the stone are: A. ( 0.4 m, 4 s ) B. ( 0.2 m, 2 s ) c. ( 0.4 m, 2 s ) D. ( 0.8 m, 2 s ) | 11 |
| 551 | A particle of mass ( 0.5 mathrm{kg} ) is executing SHM along a straight line. Its path length is ( 10 mathrm{cm} ) and time period is 8s. TE when its phase angle is ( frac{pi}{6} ) radian is B. 0.964×10-4 J c. ( 3.856 times 10^{-4} ) j D. 4.324×10-4 J | 11 |
| 552 | The equation of motion for an oscillating particle is given by ( boldsymbol{x}=boldsymbol{3} sin (boldsymbol{4} boldsymbol{pi} boldsymbol{t})+boldsymbol{4} cos (boldsymbol{4} boldsymbol{pi} boldsymbol{t}), ) where ( boldsymbol{x} ) is in mm and t is in second This question has multiple correct options A. The motion is simple harmonic B. The period of oscillation is 0.5 s c. The amplitude of oscillation is ( 5 mathrm{mm} ) D. The particle starts its motion from the equilibrium | 11 |
| 553 | n figure, ( boldsymbol{k}=mathbf{1 0 0} boldsymbol{N} / boldsymbol{m}, boldsymbol{M}= ) 1 kgand ( F=10 N . ) Write the potential energy of the spring when the block is at the right extreme during simple harmonic motion If your answer is ( x ) write the value of ( 4 x ) | 11 |
| 554 | Two strings ( A ) and ( B ) have lengths ( i_{A} ) and ( l_{B} ) and carry pendulum of masses ( M_{A} ) and ( M_{B} ) at their lower ends the upper ends being supported by rigid supports. If ( n_{A} ) and ( n_{B} ) are their frequencies of their oscillations and ( boldsymbol{n}_{boldsymbol{A}}=boldsymbol{2} boldsymbol{n}_{boldsymbol{B}}, ) then : A ( . l_{A}=4 l_{B} ), regardless of masses B . ( l_{B}=4 l_{4}, ) regardless of masses C . ( M_{A}=2 M_{B}, l_{A}=2 l_{B} ) D. ( M_{B}=2 M_{A}, l_{B}=2 l_{A} ) | 11 |
| 555 | The displacement of a particle along the ( x ) -axis is given by ( x=a sin ^{2} t . ) The motion of the particle corresponds to A. simple harmonic motion of frequency ( omega / pi ) B. simple harmonic motion of frequency 3 ( omega / 2 pi ) c. non simple harmonic motion D. simple harmonic motion of frequency ( omega / 2 pi ) | 11 |
| 556 | A uniform thin ring of radius ( boldsymbol{R} ) and mass ( m ) suspended in a vertical plane from a point in its circumference its time period of oscillation is A ( cdot 2 pi sqrt{frac{2 R}{g}} ) В ( cdot 2 pi sqrt{frac{3 R}{2 g}} ) c. ( frac{pi}{2} sqrt{frac{R}{g}} ) D. ( pi sqrt{frac{R}{2 g}} ) | 11 |
| 557 | Derive an expression for the velocity of a particle performing linear SHM using differential equation. Hence, find the expression for the maximum velocity. | 11 |
| 558 | The displacement of a particle performing linear S.H.M is given by ( boldsymbol{x}= ) ( 6 sin (3 pi t-5 pi / 6) ) metre. Find the time at which the particle reaches the extreme position towards the left: A. ( 5 / 9 ) secs B . 4/9 secs c. ( 1 / 9 ) secs D. 7/9 secs | 11 |
| 559 | Column I describe some situations in which a small object moves. Column I describes some characteristics of these motions. Match the situation in Column with the characteristics in Column I and indicate your answer by darkening appropriate bubbles in the 44 matrix given in the ORS | 11 |
| 560 | For a particle executing simple harmonic motion, the displacement x is given by ( boldsymbol{x}= ) Acoswt. Identify the graph which represents the variation of potential energy (U) as a function of time ( t ) and displacement ( x ) A . ।, II B. ॥, ॥॥ ( c cdot 1,1 v ) D. II, IV | 11 |
| 561 | A particle of mass ( m ) is executing S.H.M If amplitude is a and frequency ( n ), the value of its force constant will be: ( mathbf{A} cdot m n^{2} ) B. ( 4 m n^{2} a^{2} ) ( mathrm{c} cdot m a^{2} ) D. ( 4 pi^{2} m n^{2} ) | 11 |
| 562 | The ratio of the maximum velocity and maximum displacement of a particle executing SHM is equal to ( A cdot n ) B. ( g ) c. ( D . ) | 11 |
| 563 | Assertion A particle is under SHM along the ( x ) axis. Its mean position is ( x=2 ) amplitude is ( A=2 ) and angular frequency ( omega . A t t=0, ) particle is at origin, then ( x-c 0 ) ordinate versus time equation of the particle will be ( x=-2 cos omega t+2 ) Reason At ( t=0 ) particle is at rest. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 11 |
| 564 | A function of time given by ( (sin omega t-cos omega t) ) represents A. simple harmonic motion B. non-periodic motion c. periodic but not simple harmonic motion D. oscillatory but not simple harmonic motion | 11 |
| 565 | A student says that he had applied a force ( F=k sqrt{x} ) on a particle moved in simple harmonic motion. He refuses to tell whether ( k ) is a constant or not. Assume that he has worked only with positive ( x ) and no other force acted on the particle. A . As ( x ) increases ( k ) increases B. AS ( x ) increases ( k ) decreases C. As ( x ) increases ( k ) remains constant D. The motion cannot be simple harmonic. | 11 |
| 566 | The differential equation of a wave is : A ( cdot frac{d^{2} y}{d t^{2}}=v^{2} frac{d^{2} y}{d x^{2}} ) B. ( frac{d^{2} y}{d x^{2}}=v^{2} frac{d^{2} y}{d t^{2}} ) c. ( frac{d^{2} y}{d x^{2}}=frac{1}{v} frac{d^{2} y}{d t^{2} y} ) D. ( frac{d^{2} y}{d x^{2}}=-v frac{d^{2} y}{d t^{2}} ) | 11 |
| 567 | The graph between velocity and position for a damped oscillation will be? A. Straight line B. Circle C . Ellipse D. Spiral | 11 |
| 568 | Which of the following figure represents damped harmonic motion | 11 |
| 569 | A body is executing ( mathrm{SHM} ), At a displacement ( x, ) its potential energy is ( E_{1} ) and at a displacement ( y, ) its potential energy is ( E_{2} ). Find its potential energy ( mathrm{E} ) at displacement ( (mathrm{x}+mathrm{y}) ) B . ( sqrt{E}=sqrt{E}_{1}+sqrt{E}_{2} ) c. none of these D. all of these | 11 |
| 570 | The ratio of maximum acceleration to maximum velocity in a simple harmonic motion ( 10 S^{-1} . A t, t=0 ) the displacement is ( 5 mathrm{m} ). What is the maximum acceleration? The initial phase is ( frac{pi}{4} ) ( mathbf{A} cdot 750 sqrt{2} m / s^{2} ) B . ( 500 sqrt{2} mathrm{m} / mathrm{s}^{2} ) c. ( 750 m / s^{2} ) D. ( 500 m / s^{2} ) | 11 |
| 571 | A particle is performing linear S.H.M. with period 6 s and amplitude ‘a’. The minimum time taken by the particle to travel between two points situated at a distance ( frac{a}{2} ) on either sides of mean position is : A . 1 s в. 1.5 s ( c cdot 3 s ) D. 6 s | 11 |
| 572 | Statement 1: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Statement 2: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator. A. Statement 1 is false, Statement 2 is true B. Statement 1 is true, Statement 2 is true; Statement 2 the correct explanation for Statement c. statement 1 is true, Statement 2 is true; Statement 2 is not the correct explanation for Statement D. Statement 1 is true, Statement 2 is false | 11 |
| 573 | If a system is displaced from its equilibrium position and released, it moves according to the equation ( ddot{boldsymbol{theta}}=-frac{boldsymbol{I}^{2}}{boldsymbol{k} boldsymbol{l}} boldsymbol{theta} ) where ( I, ) k and I are constants. It will oscillate with a frequency: ( ^{A} cdot sqrt{frac{I^{2}}{k l}} ) в. ( 2 pi sqrt{frac{k l}{I^{2}}} ) ( ^{mathrm{c}} cdot frac{1}{2 pi} sqrt{frac{I^{2}}{k l}} ) D. ( frac{1}{2 pi} sqrt{frac{k l}{I^{2}}} ) | 11 |
| 574 | The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are ( 2 m / s ) and ( 4 m / s^{2} . ) The angular velocity will be: A. ( 3 r a d / s ) B. ( 0.5 r a d / s ) c. ( 1 r a d / s ) D. ( 2 r a d / s ) | 11 |
| 575 | A particle executes simple harmonic oscillation with an amplitude ( a ). The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is A ( cdot frac{T}{4} ) в. ( frac{T}{8} ) c. ( frac{T}{12} ) D. ( frac{T}{2} ) | 11 |
| 576 | For periodic motion of small amplitude ( A, ) the time period ( T ) of this particle is proportional to ( mathbf{A} cdot A sqrt{frac{m}{alpha}} ) B. ( frac{1}{A} sqrt{frac{m}{alpha}} ) c. ( A sqrt{frac{alpha}{m}} ) D. ( A sqrt{frac{2 alpha}{m}} ) | 11 |
| 577 | Select the appropriate statement from the below: A. The magnitude of maximum acceleration is ( A omega ) B. The magnitude of maximum acceleration is ( A omega^{2} ) c. The magnitude of minimum acceleration is ( A omega ) D. The magnitude of minimum acceleration is ( A omega / 2 ) | 11 |
| 578 | Ratio of kinetic energy at mean position to potential energy at ( A / 2 ) of a particle performing SHM. A . 2: 1 B . 4: 1 c. 8: 1 D. 1: 1 | 11 |
| 579 | If the time period of revolution of earth is 365 days, then, the frequency of revolution of earth is A ( cdot 3.17 times 10^{-6} mathrm{Hz} ) В. ( 3.17 times 10^{-8} mathrm{Hz} ) c. ( 3.17 times 10^{-4} mathrm{Hz} ) D. ( 3.17 times 10^{-7} mathrm{Hz} ) | 11 |
| 580 | The differential equation of a particle executing SHM along y-axis is A ( cdot frac{d^{2} y}{d t^{2}}+omega^{2} y=0 ) B. ( frac{d^{2} y}{d t^{2}}-omega^{2} y=0 ) ( ^{text {C } cdot frac{d^{2} x}{d t^{2}}+omega^{2} x=0} ) D. ( frac{d^{2} x}{d t^{2}}-omega^{2} x=0 ) | 11 |
| 581 | An oscillator is producing FM waves of frequency ( 2 mathrm{kHz} ) with avariation of 10kHz. The modulating index=? A . 0.20 B. 5.0 c. 0.67 D. 1.5 | 11 |
| 582 | State whether true or false: The particle is moving simple harmonically A. True B. False | 11 |
| 583 | Show that motion of bob of pendulum with small amplitude is linear ( S . H . M ) Hence obtained an expression for its period. What are the factors on which its period depends? | 11 |
| 584 | Dampers are found on bridges A. to allow natural oscillations to occur B. to prevent them from swaying due to wind c. to prevent resonance of frequencies. D. None of these. | 11 |
| 585 | A simple pendulum of length ( 4 mathrm{m} ) is taken to a height ( R ) (radius of the earth) from the earth’s surface.The time period of small oscillations of the pendulum is ( left(boldsymbol{g}_{text {surface}}=boldsymbol{pi}^{2} boldsymbol{m} boldsymbol{s}^{-2}right) ) A . 2 s B. 4 s ( c cdot 8 s ) D. 16 s | 11 |
| 586 | Amplitude of oscillation of a particle that executes S.H.M. is 2 cm. Its displacement from its mean position in a time equal to ( 1 / 6 ) th of its time period is A ( cdot sqrt{2} mathrm{cm} ) B. ( sqrt{3} mathrm{cm} ) c. ( 1 / sqrt{2} mathrm{cm} ) D. ( 1 / sqrt{3} mathrm{cm} ) | 11 |
| 587 | ‘The motion of a particle with a restoring force gives oscillatory motion”. Which of the following forces can be a restoring force for the oscillatory motion. A. Frictional force B. A constant force c. A time varying force opposite to direction of motion D. Force due to a spring | 11 |
| 588 | A plane progressive wave is given by ( boldsymbol{y}=25 cos (2 pi t-pi x) ) Then the amplitude and frequency are respectively A . 25,100 B. 25, ( c .25 .2 ) D. ( 50 pi, 2 ) | 11 |
| 589 | A magnet is suspended in such a way that it oscillate it horizontal plane. It make 20 oscillation per minute at a place were dip angle is 30 and 15 oscillation per minute t a place were dip angle is ( 60 . ) the ratio of total magnetic field at the two place is A . 15 oscillations/minute в. ( 1.7 sqrt{3} ) c. 18 oscillations/minute D. zero | 11 |
| 590 | A particle of mass ( m ) is attached to three springs ( A, B ) and ( C ) of equal force constants ( k ) as shown in figure. If the particle is pushed slightly against the spring ( C ) and released, the time period of oscillation is A ( cdot 2 pi sqrt{frac{m}{k}} ) В ( cdot 2 pi sqrt{frac{m}{2 k}} ) c. ( 2 pi sqrt{frac{m}{3 k}} ) D. ( 2 pi sqrt{frac{3 m}{k}} ) | 11 |
| 591 | A particle executes SHM with amplitudes ( 25 mathrm{cm} ) and period 3 seconds. The minimum time required for it to move between two points 12.5 ( mathrm{cm} ) on either side of the mean position is A. ( 0.25 mathrm{sec} ) B . 0.50 sec c. ( 0.75 mathrm{sec} ) D. ( 1.00 mathrm{sec} ) | 11 |
| 592 | Find the frequency of osculation of an underdamped harmonic oscillator of mass ( m ) in terms of its natural frequency ( omega_{0} ) and damping constant ( gamma ) (where – ( boldsymbol{v} boldsymbol{y} ) is the damping force, ( boldsymbol{v} ) being the velocity) | 11 |
| 593 | A particle executes simple harmonic motion between ( boldsymbol{x}=-boldsymbol{A} ) and ( boldsymbol{x}=+boldsymbol{A} ) The time taken for it to go from 0 to ( A / 2 ) is ( T_{1} ) and to ( operatorname{gofrom} A / 2 ) to ( A ) is ( T_{2} ). Then A ( cdot T_{1}T_{2} ) c. ( T_{1}=T_{2} ) D. ( T_{1}=2 T_{2} ) | 11 |
| 594 | A particle starts ( S . H . M . ) from the mean position. Its amplitude is ( A ) and time period is ( T . ) At the time when its speed is half of the maximum speed, its displacement ( y ) is : A ( cdot frac{A}{2} ) B. ( frac{A}{sqrt{2}} ) c. ( frac{A sqrt{3}}{2} ) D. ( frac{2 A}{sqrt{3}} ) | 11 |
| 595 | Part of a simple harmonic motion is graphed in the figure, where ( y ) is the displacement from the mean position. The correct equation describing this S.H.M. is | 11 |
| 596 | The diagram shows the displacementtime graph for a vibrating body. Name the type of vibration produced by the vibrating body : | 11 |
| 597 | The displacement of a particle is given by ( vec{r}=A(vec{i} cos omega t+vec{jmath} sin omega t) . ) The motion of the particle is: A. simple harmonic B. On a straight line c. on a circle D. With constant acceleration | 11 |
| 598 | A particle moves such that its acceleration is given by ( boldsymbol{a}=-boldsymbol{beta}(boldsymbol{x}-boldsymbol{2}) ) Here ( beta ) is a positive constant and ( x ) is the position from origin. Time period of oscillations is A. ( 2 pi sqrt{beta} ) в. ( 2 pi sqrt{frac{1}{beta}} ) ( mathbf{c} cdot 2 pi sqrt{beta}+2 ) D. ( 2 pi sqrt{frac{1}{beta+2}} ) | 11 |
| 599 | A simple pendulum completes 40 oscillation in a minute.Find its (a) frequency (b) time period. | 11 |
| 600 | Find the time period of the motion of the particle shown in figure. Neglect the small effect of the bend near the bottom. | 11 |
| 601 | If a body mass ( 36 g m ) moves with ( mathrm{S}, mathrm{H}, mathrm{M} ) of amplitude ( A=13 ) and period ( T= ) 12 ( sec ) At a time ( t=0 ) the displacement is ( x=+13 c m . ) The shortest time of passage from ( boldsymbol{x}=+mathbf{6 . 5} mathrm{cm} ) to ( boldsymbol{x}=-mathbf{6 . 5} ) is ( A cdot 4 sec ) B. 2 sec ( c cdot 6 sec ) D. 3 sec | 11 |
| 602 | In the figure, the vertical sections of the string are long. ( A ) is released from rest from the position shown. Then A. The system will remain in equilibrium B. The central block will move down continously C. The central block will undergo simple harmonic motion D. The central block will undergo periodic motion but not simple harmonic motion. | 11 |
| 603 | The time period of a particle in simple harmonic motion is 8 s. At ( t=0 ), it is at the mean position. The ratio of the distances travelled by it in the first and second seconds is: A ( cdot frac{1}{2} ) в. ( frac{1}{sqrt{2}} ) c. ( frac{1}{sqrt{2}-1} ) D. ( frac{1}{sqrt{3}} ) | 11 |
| 604 | A particle executing S.H.M. completes a distance (taking friction as negligible) in one complete one time period. A. Four times the amplitude B. Two times the amplitude c. one times the amplitude D. Eight times the amplitude | 11 |
| 605 | A particle of mass ( 10 g ) is executing simple harmonic motion with an amplitude of ( 0.5 m ) and periodic time of ( left(frac{pi}{5}right) s . ) The maximum value of the force acting on the particle is: A . ( 25 ~ N ) B. ( 5 N ) c. ( 2.5 N ) D. ( 0.5 N ) | 11 |
| 606 | The number of vibrations made by a vibrating body in one second is called its : A. wavelength B. time period c. amplitude D. frequency | 11 |
| 607 | A student performs an experiment for determination of ( left(g=frac{4 pi^{2} l}{T^{2}}right), mid=1 mathrm{m} ) and he commits an error of ( Delta l ) For T he takes the time of ( n ) oscillations with the stop watch of least count ( Delta boldsymbol{T} ) and he commits a human error of 0.1 s. For which of the following data, the measurement of ( g ) will be most accurate? A. ( Delta L=0.5, Delta T=0.1, n=20 ) B . ( Delta L=0.5, Delta T=0.1, n=50 ) c. ( Delta L=0.5, Delta T=0.01, n=20 ) D. ( Delta L=0.5, Delta T=0.05, n=50 ) | 11 |
| 608 | A ball of radius a is made to oscillate in a bowl of radius b then time period of vibration of ball is ( (boldsymbol{b}>boldsymbol{a}) ) A ( cdot 2 pi sqrt{frac{a}{g}} ) В ( cdot 2 pi sqrt{frac{b-a}{g}} ) ( ^{mathrm{c}} cdot 2 pi sqrt{frac{b}{g}} ) D. ( 2 pi sqrt{frac{b+a}{q}} ) | 11 |
| 609 | A particle of mass ( m ) is executing S.H.M. of time period ( T, ) and amplitude ( a_{0} . ) The force on particle at the mean position is: A ( cdot frac{4 pi^{2} m}{T^{2}} a_{0} ) ( ^{text {В } cdot frac{2 pi^{2} m}{T^{2}} a_{0}} ) c. zero D. ( frac{pi^{2} m a_{0}}{T^{2}} ) | 11 |
| 610 | What are damped vibrations? How do they differ from free vibrations ? Give one example of each. | 11 |
| 611 | A boy is executing Simple Harmonic Motion. At a displacement ( x ), its potential energy is ( E_{1} ) and at a displacement ( y, ) its potential energy is ( boldsymbol{E}_{2} . ) The potential energy ( boldsymbol{E} ) at displacement ( (boldsymbol{x}+boldsymbol{y}) ) is: A ( cdot sqrt{E}=sqrt{E_{1}}-sqrt{E_{2}} ) B. ( sqrt{E}=sqrt{E_{1}}+sqrt{E_{2}} ) c. ( E=E_{1}+E_{2} ) D. ( E=E_{1}-E_{2} ) | 11 |
| 612 | The two particles are at minimum distance from each other after time, ( t= ) ( mathcal{S} ) A . в. 2. ( c .3 ) D. all of these | 11 |
| 613 | In forced oscillation of a particle the amplitude is maximum for a frequency ( omega_{1} ) of force, while the energy is maximum for a frequency ( omega_{2} ) of the force, then: A. ( omega_{1}=omega_{2} ) В. ( omega_{1}>omega_{2} ) C. ( omega_{1}omega_{2} ) when damping is large D. ( omega_{1}<omega_{2} ) | 11 |
| 614 | The oscillators that can be described in terms of sine or cosine functions are called A. simple harmonic B. natural c. sympathetic D. free | 11 |
| 615 | Displacement-time graph depicting an oscillatory motion is A. cos curve B. sine curve c. tangent curve D. straight line | 11 |
| 616 | The ratio of the angular speed of minutes hand and hour hand of a watch is: A . 6: 1 B. 12: 1 c. 1: 6 D. 1: 12 | 11 |
| 617 | A particle executes ( boldsymbol{S} boldsymbol{H} boldsymbol{M} ) of amplitude ( 25 c m ) and time period ( 3 s ) What is the minimum time required for the particle to move between two points ( 12.5 mathrm{cm} ) on either side of the mean position? A . ( 0.5 s ) B. ( 1.0 s ) c. ( 1.5 s ) D. 2.0 | 11 |
| 618 | Assertion In SHM let ( x ) be the maximum speed, ( y ) the frequency of oscillation and ( z ) the maximum acceleration then ( frac{x y}{z} ) is a constant quantity. Reason This is because ( frac{x y}{z} ) becomes a dimensionless quantity A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 11 |
| 619 | The frequency of a sound wave is ( 256 mathrm{Hz} ) and its wavelength is ( 1.2 mathrm{m} ). Calculate its wave velocity. | 11 |
| 620 | A particle performing SHM has a period of ( 6 s ) and amplitude of ( 8 mathrm{cm} . ) The particle starts from the mean position and moves towards the positive extremity. At what time will it have the maximum amplitude A . ( 6 s ) B. 4.5s ( c cdot 1.5 s ) D. 3 s | 11 |
| 621 | A particle of mass ( mathrm{m} ) is executing oscillations about the origin on the ( x ) axis. Its potential energy is ( V(x)= ) ( k|x|^{3} ) where ( k ) is a positive constant. If the amplitude of oscillation is a, them its time period T is? A. proportional to ( frac{1}{sqrt{a}} ) B. Independent of a c. Proportional to ( sqrt{a} ) D. Proportional to ( a^{3 / 2} / 2 ) | 11 |
| 622 | The time taken by the bob of a pendulum to complete one oscillation is called its A. time period B. amplitude c. mass D. frequency | 11 |
| 623 | The slope of kinetic energy vs. displacement curve of a particle in motion is: A. inversely proportional to acceleration of the particle. B. directly proportional to acceleration of the particle. c. equal to acceleration of particle. D. none of the above | 11 |
| 624 | In the figure shown, the springs are connected to the rod at one end and at the midpoint. The rod is hinged at its lower end. Rotational SHM of the rod (Mass ( boldsymbol{m} ), length ( boldsymbol{L} ) ) will occur only if ( mathbf{A} cdot k>m g / 3 L ) В. ( k>2 m g / 3 L ) c. ( k>2 m g / 5 L ) D. ( k>0 ) | 11 |
| 625 | For a body in S.H.M. the velocity is given by the relation ( boldsymbol{v}=sqrt{144-16 x^{2}} boldsymbol{m} / boldsymbol{s} ) The maximum acceleration is A ( cdot 12 m / s^{2} ) в. ( 16 m / s^{2} ) ( mathrm{c} cdot 36 mathrm{m} / mathrm{s}^{2} ) D. ( 48 mathrm{m} / mathrm{s}^{2} ) | 11 |
| 626 | Two simple harmonic motions are represented by the equations. ( boldsymbol{y}_{1}=mathbf{1 0} sin frac{boldsymbol{pi}}{boldsymbol{4}}(1 mathbf{2} boldsymbol{t}+mathbf{1}), boldsymbol{y}_{2}= ) ( mathbf{5}(((sin 3 p t+sqrt{3} cos 3 p t) ) The ratio of their amplitudes is A . 1: 1 B. 1: 2 c. 3: 2 D. 2: 3 | 11 |
| 627 | The displacement time graph of a particle executing SHM is shown in figure. Which of the following statements is/are true? This question has multiple correct options A. The velocity is maximum at t=T/2 B. The acceleration is maximum at t= c. The force is zero at t=37/4 D. The kinetic energy equals the total oscillation energy at ( t=T / 2 ) | 11 |
| 628 | How much time the man has to board the boat comfortably during each cycle of up and down motion? A . 0.585 s в. 1.17 s ( c cdot 2.33 mathrm{s} ) D. 0.293 s | 11 |
| 629 | A hydrogen atom has mass ( 1.68 times ) ( 10^{-27} mathrm{kg} . ) When attached to a certain massive molecule it oscillates with a frequency ( 10^{14} mathrm{Hz} ) and with an amplitude ( 10^{-9} mathrm{cm} . ) Find the force acting on the hydrogen atom. B . ( 3.31 times 10^{-9} ) N c. ( 4.42 times 10^{-9} ) N D. ( 6.63 times 10^{-9} ) N | 11 |
| 630 | On suspending a mass M from a spring of force constant K,frequency of vibration ( f ) is obtained If a second spring as shown in the figure. is arranged then the frequency will be : A ( . f sqrt{2} ) B. ( f / sqrt{2} ) ( c . ) २ D. | 11 |
| 631 | The amplitude of a particle executing SHM about O is 10 cm. Then: This question has multiple correct options A. When the K.E. is 0.64 of its max. K.E. its displacement is ( 6 mathrm{cm} ) from 0 . B. When the displacement is 5 cm from 0 its K.E. is 0.75 of its max.P.E. C. Its total energy at any point is equal to its maximum K.E. D. Its velocity is half the maximum velocity when its displacement is half the maximum displacement. | 11 |
| 632 | The oscillations represented by curve 1 in the graph are expressed by equation ( x= ) Asinwt. The equation for the oscillations represented by curve 2 is expressed as: A. ( x=2 A sin (omega t-pi / 2) ) B. ( x=2 A sin (omega t+pi / 2) ) c. ( x=-2 operatorname{Asin}(omega t-pi / 2) ) D. ( x=operatorname{Asin}(omega t-pi / 2) ) | 11 |
| 633 | A particle performing SHM with a frequency of ( 5 mathrm{Hz} ) and amplitude ( 2 mathrm{cm} ) is initially in the positive extreme position. The equation for its displacement is A. ( x=0.02 sin 10 pi t mathrm{m} ) B. ( x=0.02 sin 5 pi t mathrm{m} ) c. ( x=0.02 cos 10 pi t mathrm{m} ) D. ( x=0.02 cos 5 pi t mathrm{m} ) | 11 |
| 634 | The displacement of a particle in S.H.M. is indicated by equation ( boldsymbol{y}= ) ( 10 sin (20 t+Omega / 3) ) where ( y ) is in metres. The value of time period of vibration will be (in seconds): A . ( 10 / pi ) в. ( pi / 10 ) c. ( 2 pi / 10 ) D. ( 10 / pi 2 ) | 11 |
| 635 | Assertion The time period of a simple pendulum is independent of its length. Reason The length of a pendulum is the distance between point of suspension and centre of the bob A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 11 |
| 636 | A particle moves under the force ( boldsymbol{F}(boldsymbol{x})=left(boldsymbol{x}^{2}-boldsymbol{6} boldsymbol{x}right) boldsymbol{N}, ) where ( boldsymbol{x} ) is in meters. Show that for small displacements from the origin the force constant in the simple harmonic motion is approximately 6 | 11 |
| 637 | A block performs simple harmonic motion with equilibrium point ( boldsymbol{x}=mathbf{0} ) Graph of acceleration of the block as a function of time is shown. Which of the following statement is correct about the block? A. Displacement from equilibrium is maximum at ( t=4 s ) B. Speed is maximum at ( t=4 s ) C. Speed is maximum at ( t=2 s ) D. Speed is maximum at ( t=3 s ) | 11 |
| 638 | Two particles are in SHM in parallel straight lines close to each other. Amplitude ( A ) and time period ( T ) of both the particles are equal. At ( t=0, ) one particle is at displacement ( boldsymbol{x}_{mathbf{1}}=+boldsymbol{A} ) and other at ( x_{2}=-frac{A}{2} ) an they are approaching towards each other, after what time they will cross each other? A. ( T / 3 ) в. ( T / 4 ) ( mathrm{c} cdot 5 T / 6 ) D. ( T / 6 ) | 11 |
| 639 | The force which tries to bring the body back to its mean position is called A. deforming force B. restoring force c. gravitational force D. buoyant force | 11 |
| 640 | A body describing SHM has a maximum acceleration of ( 8 pi mathrm{m} / mathrm{s}^{2} ) and a maximum speed of ( 1.6 mathrm{m} / mathrm{s} ). Find the period ( T ) and the amplitude ( A ) | 11 |
| 641 | Give two practical applications of simple harmonic motion. | 11 |
| 642 | represents the position-time graph for a spring-mass system oscillating with simple harmonic motion. The colored, dashed graphs represent shapes of possible velocity time graphs for the same motion. The vertical axis stands for position or | 11 |
| 643 | The potential energy of a particle varies with distance ( x ) as shown in the graph. The force acting on the particle is zero at: ( A ). B ( c . ) В and ( c ) D. A and | 11 |
| 644 | The potential energy of a particle is directly proportional to its linear displacement from its mean position. Then, the particle performs a A. Retarding straight line motion B. Damped SHM c. Linear SHM D. Angular SHM | 11 |
| 645 | The period of small oscillations is A ( cdot pi a sqrt{frac{m}{C}} ) В. ( pi a sqrt{frac{2 m a}{c}} ) C ( cdot 2 pi sqrt{frac{2 m a^{3}}{C}} ) D. ( 2 pi a sqrt{frac{m a}{C}} ) | 11 |
| 646 | The negative sign in the equation for ( f_{r} ) says that A. the direction of force is sometimes opposite to the direction of body’s motion B. the force is in the same direction to the body’s motion c. the force is in the opposite direction to the body’s motion D. None of these | 11 |
| 647 | The amplitude and the periodic time of a SHM are ( 5 c m ) and ( 6 s ) respectively. At a distance of ( 2.5 mathrm{cm} ) away from the mean position, the phase will be ( ^{A} cdot frac{pi}{3} ) в. c. D. ( frac{5 pi}{12} ) | 11 |
| 648 | A particle of mass ( m ) is moving along the X-axis under the potential ( U(x)= ) ( frac{k x^{2}}{2}+lambda ) where ( k ) and ( lambda ) are positive constants of appropriate dimensions. The particle is slightly displaced from its equilibrium position. The particle oscillates with the angular frequency ( (omega) ) given by A ( cdot_{3} frac{k}{m} ) в. ( 3 frac{m}{k} ) c. ( sqrt{frac{k}{m}} ) D. ( sqrt{3 frac{m}{k}} ) E ( cdot sqrt{3 frac{k}{m}} ) | 11 |
| 649 | A simple harmonic motion has an amplitude ( A ) and time period ( T . ) Find the time required by it to travel directly from ( boldsymbol{x}=boldsymbol{A} ) to ( boldsymbol{x}=boldsymbol{A} / mathbf{2} ) | 11 |
| 650 | A plot of potential energy v/s kinetic energy of a particle executing SHM gives a straight line A. passing through the origin B. with positive slope c. with intercepts on both the axes D. parallel to either of the axes | 11 |
| 651 | The period of oscillation of a simple pendulum of length lis given by ( boldsymbol{T}= ) ( 2 pi sqrt{l / g} . ) The length lis about ( 10 mathrm{cm} ) and is known to ( 1 mathrm{mm} ) accuracy. The period of oscillation is about 0.5 s. The time of 100 oscillations has been measured with a stop watch of 1 s resolution. Find the percentage error in the determination of ( mathfrak{g} ) | 11 |
| 652 | A particle executes simple harmonic motion with a frequency ( v ). The frequency with which the kinetic energy oscillates is ( mathbf{A} cdot v / 2 ) B. ( v ) c. ( 2 v ) D. Zero | 11 |
| 653 | The amplitude of a damped oscillator becomes ( left(frac{1}{3}right) r d ) in ( 2 s . ) If its amplitude ( operatorname{after} 6 s operatorname{in} frac{1}{n} ) times the original amplitude, the value of ( n ) is ( A cdot 3^{2} ) B. ( 3 sqrt{2} ) ( c cdot 3^{3} ) D. ( 2^{3} ) | 11 |
| 654 | If a simple harmonic motion is represented by ( frac{d^{2} x}{d t^{2}}+alpha x=0, ) its time period is then A. ( 2 pi sqrt{alpha} ). . ( sqrt{sqrt{alpha}} ). ( alpha ). в. ( 2 pi alpha ) c. ( frac{2 pi}{sqrt{alpha}} ) D. ( frac{2 pi}{alpha} ) | 11 |
| 655 | By suspending a mass of ( 0.50 mathrm{kg} ), a spring is stretched by ( 8.20 mathrm{m} . ) If a mass of ( 0.25 mathrm{kg} ) is suspended, then its period of oscillation will be: (Take ( g=10 m s^{-2} ) ) A . 0.137 s B. 0.328 s c. 0.628 s D. 1.000 | 11 |
| 656 | Which of the following equations represents a particle performing simple harmonic motion A ( cdot x=3 ) B. ( x=4 sin 2 t ) ( c cdot x=1 / t ) D. ( x=3 log 2 t ) | 11 |
| 657 | The mean value of velocity vector projection average over ( 3 / 8 ) th of a period after the start is A ( cdot frac{2 sqrt{2} A omega}{pi} ) B. ( frac{2 A omega}{3 sqrt{2 pi}} ) c. ( frac{A omega}{3 pi} ) D. ( frac{2 sqrt{2} A omega}{3 pi} ) | 11 |
| 658 | The acceleration (a) of SHM at mean position is : A. zero B. ( propto x ) ( c cdot propto x^{2} ) D. None of these | 11 |
| 659 | A rigid body rotates about a fixed axis with variable angular velocity equal to ( (a b t) ) at ( t ) where a and ( b ) are constants. The angle through which it rotates before it comes to rest is A ( cdot frac{a^{2}}{b} ) в. ( frac{a^{2}}{2 b} ) c. ( frac{a^{2}}{4 b} ) D. ( frac{a^{2}}{2 b^{2}} ) | 11 |
| 660 | A particle executes ( S H M ) of period ( 1.2 mathrm{sec} ) and amplitude ( 8 mathrm{cm} . ) Find the time it takes to travel ( 3 mathrm{cm} ) from the positive extremity or its oscillation. ( left[cos ^{-1}(5 / 8)=0.9 r a dright] ) A. 0.28 sec B. 0.32 sec c. 0.17 sec D. 0.42 sec | 11 |
| 661 | In a simple harmonic motion A. the maximum potential energy equals the maximum kinetic energy B. the minimum potential energy equals the minimum kinetic energy C. the minimum potential energy equals the maximum kinetic energy D. the maximum potential energy equals the minimum kinetic energy | 11 |
| 662 | A particle executes simple harmonic motion with a frequency ( f ). The frequency with which the potential energy oscillates is: ( A cdot f ) B. ( frac{f}{2} ) c. ( 2 f ) D. zero | 11 |
| 663 | Two wave are represented by equation ( boldsymbol{y}_{1}=boldsymbol{a} sin omega t ) and ( boldsymbol{y}_{2}=boldsymbol{a} cos omega boldsymbol{t} . ) Then the first wave: A. leads the second by ( pi ) B. lags the second by ( pi ) c. leads the second by ( frac{pi}{2} ) D. lags the second by ( frac{pi}{2} ) | 11 |
| 664 | Match List – I (Event) with List – II (Order of the time interval for happening of the event) and select the correct option from the options given below the lists. List – 1 (a) Rotation period of (i) earth Revolution period of (ii) (b) earth ( (c) ) Period of a light (iii) wave ( (d) ) Period of a sound iv wave A ( cdot(a)-(i),(b)-(i i),(c)-(i i i),(d)-(i v) ) B. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) ( c cdot(a)-(i),(b)-(i i),(c)-(i v),(d)-(i i i) ) D. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv) | 11 |
| 665 | A particle executing SHM has a maximum speed of ( 0.5 m s^{-1} ) and maximum acceleration of ( 1 m s^{-2} ) The angular frequency of oscillation is: A ( cdot 2 r a d s^{-1} ) B . 0.5 rads( ^{-1} ) C ( .2 pi r a d s^{-1} ) D. ( 0.5 pi r a d s^{-1} ) | 11 |
| 666 | Mention any two characteristics of SHM (Simple Harmonic Motion). | 11 |
| 667 | The phase angle between the projections of uniform circular motion on two mutually perpendicular diameter is A . ( pi ) в. ( 3 pi / 4 ) c. ( pi / 2 ) D. zero | 11 |
| 668 | The displacement (in ( mathrm{m} ) ) of a particle of mass 100 gram from its equilibrium position is given by ( boldsymbol{y}=mathbf{0 . 0 1} sin 2 pi(t+ ) 0.4). Select the correct option(s): This question has multiple correct options A. The time period of motion is 1 sec. B. The time period of motion is ( frac{1}{7.5} ) sec c. The maximum acceleration of the particle is ( 0.04 pi^{2} m / s^{2} ) D. The force acting on the particle is zero when the displacement is 0.05m | 11 |
| 669 | A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is: A. ( T / 2 ) в. ( T / 4 ) ( c cdot T / 8 ) D. T/12 | 11 |
| 670 | ( x-t ) equation of a particle in SHM is, ( boldsymbol{x}=mathbf{4}+mathbf{6} sin boldsymbol{pi} boldsymbol{t} ) Match the following tables corresponding to time taken in moving from: | 11 |
| 671 | A toy train rotates about a circle of radius ( 50 mathrm{cms}, 10 ) times in 2 minutes. What is the frequency of the train A. 5 per second B. 12 per second c. ( 1 / 12 ) per second D. ( 1 / 5 ) per second | 11 |
| 672 | Assertion In case of oscillatory motion the average speed for any time interval is always greater than or equal to its average velocity Reason Distance travelled by a particle cannot be less than its displacement. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 11 |
| 673 | Which of the following functions represent a simple harmonic motion? ( mathbf{A} cdot sin omega t-cos omega t ) B. ( sin ^{2} omega t ) ( mathbf{c} cdot sin omega t+sin ^{2} omega t ) D. ( sin omega t-sin ^{2} omega t ) | 11 |
| 674 | If a simple harmonic motion is represent by ( frac{d^{2} x}{d t^{2}}+a x=0, ) its time period is A ( cdot frac{2 pi}{alpha} ) в. ( frac{2 pi}{sqrt{alpha}} ) ( c cdot 2 pi alpha ) D. ( 2 pi sqrt{alpha} ) | 11 |
| 675 | The period of pendulum depends upon A. mass B. length c. amplitude D. energy | 11 |
| 676 | One end of an ideal spring is fixed to a wall at origin ( O ) and axis of spring is parallel to x-axis. A block of mass ( boldsymbol{m}= ) 1 kg is attached to free end of the spring and it is performing SHM. Equation of position of the block in coordinate system shown in figure is ( boldsymbol{x}=mathbf{1 0}+mathbf{3} sin (mathbf{1 0} boldsymbol{t}) . ) Here, ( boldsymbol{t} ) is in second and ( x ) in cm. Another block of mass ( M=3 k g, ) moving towards the origin with velocity ( 30 mathrm{cm} / mathrm{s} ) collides with the block performing SHM at ( t=0 ) and gets stuck to it. Calculate new amplitude of oscillations | 11 |
| 677 | In a damped harmonic oscillator, periodic oscillations have amplitude. A. Gradually increasing B. Suddenly increasing c. suddenly decreasing D. Gradually decreasing | 11 |
| 678 | Calculate the velocity of the bob of a simple pendulum at its mean position if it is able to rise to a vertical height of ( 10 mathrm{cm} . ) Given: ( g=980 mathrm{cms}^{-2} ) | 11 |
| 679 | State the differential equation of linear simple harmonic motion. Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear simple harmonic motion. A body cools from ( 80^{circ} mathrm{C} ) to ( 70^{circ} mathrm{C} ) in 5 | 11 |
| 680 | Assertion For a particle performing SHM, its speed decreases as it goes away from the mean position. Reason In SHM, the acceleration is always opposite to the velocity of the particle. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 11 |
| 681 | The motion of a particle is given by ( boldsymbol{x}= ) ( a sin omega t+b cos omega t . ) The motion of the particle is: A. Not simple harmonic B. Simple harmonic with amplitude ( (A-B) / 2 ) C. Simple harmonic with amplitude ( (A+B) / 2 ) D. Simple harmonic with amplitude ( sqrt{A^{2}+B^{2}} ) | 11 |
| 682 | The displacement ( y ) of a particle executing periodic motion is given by ( y=4 cos ^{2}left(frac{1}{2} tright) sin (1000 t) . ) This expression may be considered to be a result of the superposition of independent harmonic motions A. two B. three c. four D. five | 11 |
| 683 | Undamped oscillations are practically impossible because A. there is always loss of energy. B. there is no force opposing friction. C. energy is not conserved in such oscillations D. None of these. | 11 |
| 684 | Assertion : In damped oscillations, the energy of the system is dissipated continuously. Reason: For the small damping, the oscillations remain approximately periodic. A. If both assertion and reason are true and reason is the correct explanation of assertion. B. If both assertion and reason are true and reason is not the correct explanation of assertion. c. If assertion is true but reason is false D. If both assertion and reason are false | 11 |
| 685 | How much energy has been lost during these four oscillations? A . 0.31 B. 0.41 c. 0.51 D. 0.39 | 11 |
| 686 | The position and velocity of a particle executing simple harmonic motion at ( t=0 ) are given by ( 3 mathrm{cm} ) and ( 8 mathrm{cm} / mathrm{s} ) respectively. If the angular frequency of the particle is ( 2 r a d / s ), then the amplitude of oscillation, in centimeters is? A . 3 B. 4 c. 5 D. 6 ( E ) | 11 |
| 687 | The direction of velocity of a pendulum bob at the right extreme position is: A. towards right B. towards north c. towards the centre D. towards left | 11 |
| 688 | The amplitude of a particle executing SHM about 0 is ( 10 mathrm{cm} . ) Then: A. when the KE is 0.64 times of its maximum KE, its displacement is 6 cm from 0 B. its speed is half the maximum speed when its displacement is half the maximum displacement ( c . ) Both (a) and (b) are correct D. Both (a) and (b) are wrong | 11 |
| 689 | A particle is acted upon by a force given by ( boldsymbol{F}=-boldsymbol{alpha} boldsymbol{x}^{3}-boldsymbol{beta} boldsymbol{x}^{4} ) where ( boldsymbol{alpha} ) and ( boldsymbol{beta} ) are positive constants. At the point ( x=0 ) the particle is A. In stable equilibrium B. In unstable equilibrium c. In neutral equilibrium D. Not in equilibrium | 11 |
| 690 | The angle between the instantaneous velocity and acceleration of a particle executing SHM is A. zero or ( pi ) В. ( pi / 2 ) c. zero D. | 11 |
| 691 | A simple pendulum is oscillating with amplitude ( A ) and angular frequency ( omega ) At displacement ( x ) from mean position, the ratio of kinetic energy to potential energy is A ( cdot frac{x^{2}}{A^{2}-x^{2}} ) B. ( frac{x^{2}-A^{2}}{x^{2}} ) c. ( frac{A^{2}-x^{2}}{x^{2}} ) D. ( frac{A-x}{x} ) | 11 |
| 692 | Two spring of force constant ( 300 mathrm{N} / mathrm{m} ) (spring ( A ) ) and ( 400 mathrm{N} / mathrm{m} ) (spring ( mathrm{B} ) ) are joined together in series. The combination is compressed by ( 8.75 mathrm{cm} ) The ratio of energy stored in ( A ) and ( B ) is ( underline{E}_{boldsymbol{B}_{boldsymbol{B}}} ) A ( cdot frac{4}{3} ) в. ( frac{16}{9} ) ( c cdot frac{3}{4} ) D. ( frac{9}{16} ) | 11 |
| 693 | For a particle performing linear S.H.M. Its average speed over one oscillation is ( (a=a m p l i t u d e text { of } S . H . M, n= ) frequency of oscillation) A . ( 2 a n ) B. 4an c. ( 6 a n ) D. ( 8 a n ) | 11 |
| 694 | The angular frequency of the damped oscillator is given by ( omega= ) ( sqrt{left(frac{k}{m}-frac{r^{2}}{4 m^{2}}right)} ) where ( k ) is the spring constant, ( m ) is the mass of the oscillator and ( r ) is the damping constant. If the ratio ( frac{r^{2}}{m k} ) is ( 8 % ), the changed in time period compared to the undamped oscillator is approximately as follows: A. Increases by ( 1 % ) B. Decreases by 1% c. Decreases by ( 8 % ) D. increases by 8% | 11 |
| 695 | The displacement of a particle in S.H. ( M . ) is indicated by equation ( y= ) ( mathbf{1 0} sin (mathbf{2 0 t}+boldsymbol{pi} / mathbf{3}) ) where ( boldsymbol{y} ) is in metres. The value of maximum velocity of the particle will be : A . ( 100 mathrm{m} / mathrm{sec} ) B . 150 m/sec. c. ( 200 mathrm{m} / mathrm{sec} ) D. 400 ( mathrm{m} / mathrm{sec} ) | 11 |
| 696 | For any SHM, amplitude is ( 6 mathrm{cm} ). If instantaneous potential energy is half the total energy then distance of particle from its mean position is A. 3 cm B. ( 4.2 mathrm{cm} ) c. ( 5.8 mathrm{cm} ) D. 6 cm | 11 |
| 697 | A horizontal rod of length L and mass ( mathrm{m} ) lying on the floor is fixed at one end and a force ( F ) is applied at an angle ( theta ) with the horizontal. The torque experienced by the rod is ( mathbf{A} cdotleft(m L^{2} alpharight) / 12 ) В. ( left(m L^{2} alpharight) / 3 ) c. ( left(m L^{2} alpharight) / 4 ) D. ( left(m L^{2} alpharight) / 2 ) | 11 |
| 698 | A particle executes SHM on a straight line. At two positions it’s velocity ( u ) and ( boldsymbol{v} ) while acceleration, ( boldsymbol{alpha} ) and ( boldsymbol{beta} ) respectively ( [beta>alpha>0] . ) the distance between the these two positions is A ( cdot frac{u^{2}+v^{2}}{alpha+beta} ) в. ( frac{u^{2}-v^{2}}{alpha+beta} ) c. ( frac{u^{2}-v^{2}}{beta-alpha} ) D. ( frac{u^{2}+v^{2}}{beta-alpha} ) | 11 |
| 699 | Q Type your question_ sornetıries consider a thingy aldacried to a horizontal spring and moving horizontally on a frictionless surface, instead of the hanging thingy that we have been looking at. Suppose that the two springs and the two thingies are identical. Think about whether these two systems are significantly different in other respects and decide which one of the following statements is true. A. The systems have different periods because their motions are aligned differently with the gravitational field. B. The hanging system has a slightly smaller period because the weight of the spring has to be accounted for. C. The hanging system has a slightly larger period because the weight of the spring has to be accounted for. D. The two systems have identical periods, no matter what the weight of the spring is. | 11 |
| 700 | Two blocks ( P ) and ( Q ) of masses ( 0.3 k g ) and 0.4 kg respectively are stuck to each other by some weak glue as shown in the figure. They hang together at the end of a spring with a spring constant ( boldsymbol{K}=mathbf{2 0 0} boldsymbol{N} boldsymbol{m}^{-1} . ) If the block ( boldsymbol{Q} ) suddenly falls free due to the failure of glue, find (i) period of SHM of block ( boldsymbol{P} ) (ii) The amplitude of its SHM (iii) The total energy of oscillation of the system. | 11 |
| 701 | For what instant of time, does potential energy becomes equal to kinetic energy for a particle executing SHM with a time period ( T ) A. ( T / 4 ) в. ( T / 8 ) c. ( T / 2 ) D. ( T ) | 11 |
| 702 | A particle executes ( S H M ) with time period ( T ) and amplitude ( A ), The maximum possible average velocity in time ( T / 4 ) is A ( cdot frac{2 A}{T} ) в. ( frac{4 A}{T} ) c. ( frac{8 A}{T} ) D. ( frac{4 sqrt{2 A}}{T} ) | 11 |
| 703 | A simple harmonic motion has an amplitude ( A ) and time period ( T ). Find the time required by it to travel directly from ( boldsymbol{x}=mathbf{0} ) to ( boldsymbol{x}=frac{boldsymbol{A}}{mathbf{2}} ) | 11 |
| 704 | The displacement – time graph for a particle executing SHM is as shown in figure Which of the following statement is correct? A ( cdot ) the velocity of the particle is maximum at ( t=frac{3}{4} T ) B. The velocity of the particle is maximum at ( t=frac{T}{2} ) C the acceleration of the particle is maximum at ( t=frac{T}{4} ) D. The acceleration of the particle is maximum at ( t=frac{3}{4} T ) | 11 |
| 705 | A particle is executing SHM with amplitude ( A ) and has maximum vekocity ( v_{0} . ) Find its speed when it is located at distance of ( frac{A}{2} ) from position. | 11 |
| 706 | Two particles ( P ) and ( Q ) describe SHM of same amplitude a and frequency along the same straight line. The maximum distance between the two particles is ( a sqrt{2} . ) The initial phase difference between the particle is ( mathbf{A} cdot pi / 3 ) в. ( pi / 2 ) c. ( pi / 6 ) D. zero | 11 |
| 707 | Which is the correct representation of the net force ( f_{n e t} ) on the mass? (k is the constant of oscillation and ( x ) is the displacement about the mean position) A ( cdot f_{n e t}=-k x+b v ) B. ( f_{n e t}=k x+b v ) C ( cdot f_{n e t}=-k x-b v ) ( mathbf{D} cdot f_{n e t}=k x-b v ) | 11 |
| 708 | How long will it be moving until it stops for the first time? A ( cdot t=2 frac{pi}{omega} ) В ( cdot t=3 frac{pi}{omega} ) c. ( t=4 frac{pi}{omega} ) D・ ( t=frac{pi}{omega} ) | 11 |
| 709 | The diagram shows the displacement- time graph for a vibrating body. Give an example of a body producing such vibrations : | 11 |
| 710 | A body executing SHM has its velocity ( 10 c m / s ) and ( 7 c m / s ) when its displacements from mean position are ( 3 c m ) and ( 4 c m ) respectively. The length of path is ( mathbf{A} cdot 10 mathrm{cm} ) B. ( 9.5 mathrm{cm} ) c. ( 4 c m ) D. ( 11.36 mathrm{cm} ) | 11 |
| 711 | A body of mass ( 1 / 4 mathrm{kg} ) is in S.H.M and its displacement is given by the relation ( boldsymbol{y}=mathbf{0 . 0 5 operatorname { s i n }}left(mathbf{2 0 t}+frac{boldsymbol{pi}}{mathbf{2}}right) mathrm{m} . ) If ( boldsymbol{t} ) is in seconds, the maximum force acting on the particle is: A . 5 N B. 2.5 N c. 10 N D. 0.25 N | 11 |
| 712 | ( ln S H M ) restoring force is ( F=-k x ) where ( k ) is force constant, ( x ) is displacement and ( a ) is amplitude of motion, then total energy depends upon A. ( k, a ) and ( m ) в. ( k, x, m ) c. ( k, a ) D. ( k, x ) | 11 |
| 713 | At resonance, the amplitude of forced oscillations is A. minimum B. maximum c. zero D. none of these | 11 |
| 714 | A particle of mass ( 0.3 mathrm{kg} ) is subjected to a force ( boldsymbol{F}=-boldsymbol{k} boldsymbol{x} ) with ( mathbf{k}=15 mathrm{N} / mathrm{m} . ) What will be its initial acceleration if it is released from a point ( x=20 mathrm{cm} ? ) | 11 |
| 715 | A comparison of the plots of the magnitude of acceleration – time graph and displacement – time graph for a particle executing SHM with angular velocity 2 rad/s reveals A. Both the curves overlap exactly with their magnitudes B. The curve appear complementary to each other C. The time periods calculated using these plots are identical D. The time period calculated in both the plots are different | 11 |
| 716 | The diagram shows the displacement- time graph for a vibrating body. Why is the amplitude of the wave gradually decreasing? | 11 |
| 717 | The figure shows the displacementtime graph of a particle executing SHM. If the time period of oscillation is ( 2 s ) then the equation of motion is given by ( boldsymbol{x}=boldsymbol{z} sin (boldsymbol{pi} boldsymbol{t}+boldsymbol{pi} / boldsymbol{6}) . ) Find ( boldsymbol{z} ) | 11 |
| 718 | The disk has a weight of ( 100 mathrm{N} ) and rolls without slipping on the horizontal surface as it oscillates about its equilibrium position. If the disk is displaced, by rolling it counterclockwise 0.4 rad, determine the equation which describes its oscillatory motion when it is released. A ( cdot theta=-0.2 cos (16.16 t) ) B ( cdot theta=0.2 cos (16.16 t) ) c. ( theta=-0.4 cos (16.16 t) ) | 11 |
| 719 | When a spring is extended by ( 2 mathrm{cm} ) energy stored is 100 J. When extended by further ( 2 mathrm{cm}, ) the energy increases by A . ( 400 J ) B. ( 300 J ) c. ( 200 J ) D. ( 100 J ) | 11 |
| 720 | A steel ball of mass ( 0.5 mathrm{kg} ) is dropped from a height of ( 4 mathrm{m} ) on to a horizontal heavy steel slap. the strile the slap and reounds to its original height. (a) calculate the impulse delivered to the ball during impact (b) if the ball is in contact with the slab for ( 0.002, ) find the average reaction force on the ball during impact. | 11 |
| 721 | A block rest on a horizontal table which is executing SHM in the horizontal with an amplitude ( a ) if the coefficient of friction is ( mu, ) then the block just start to slip when the frequency of oscillation is A ( cdot frac{1}{2 pi} sqrt{frac{mu g}{a}} ) В ( cdot 2 pi sqrt{frac{a}{mu g}} ) ( ^{mathrm{C}} cdot frac{1}{2 pi} sqrt{frac{a}{mu g}} ) D. ( sqrt{frac{a}{mu g}} ) | 11 |
| 722 | A mass of ( 5 mathrm{kg} ) is suspended from a spring of stiffness ( 46 mathrm{kN} / mathrm{m} . ) A dashpot is fitted between the mass and the support with a damping ratio of 0.3 Calculate the damped frequency. A. ( 14.56 mathrm{Hz} ) B. 14.28 Hz c. ( 14.42 mathrm{Hz} ) D. 14.14 нz | 11 |
| 723 | The motion of a pendulum is an example of : A. translatory motion B. rotational motion c. oscillatotry motion D. curvilinear motion | 11 |
| 724 | Which of the following expressions does not represent simple harmonic motion? A ( . x= ) Acoswt ( + ) Bsinwt В. ( F=sqrt{k y} ) c. ( F=k y ) D. none of these | 11 |
| 725 | Which of the following functions represents a simple harmonic oscillation? A. sinwt – coswt B. ( sin ^{2} omega t ) c. sinwt ( +sin 2 omega t ) D. ( sin omega t-sin 2 omega t ) | 11 |
| 726 | A bob is suspended by a string of length I. The minimum horizontal velocity imparted to the ball for reaching it to the height of suspension is ( mathbf{A} cdot sqrt{l / g} ) B . ( sqrt{2 g l} ) c. ( sqrt{g / l} ) D. ( 2 sqrt{g l} ) | 11 |
| 727 | A particle execute ( S H M ) from the mean position. its amplitude is ( A ) its time period ( T . ) At what displacement its speed is half of its maximum speed? A ( cdot frac{sqrt{3} mathrm{A}}{2} ) B. ( frac{sqrt{2}}{3} mathrm{A} ) ( c cdot frac{2 A}{sqrt{3}} ) ( D cdot frac{3 A}{sqrt{2}} ) | 11 |
| 728 | A particle is executing SHM with time period T. Starting from mean position, time taken by it to complete ( 5 / 8 ) oscillations, is | 11 |
| 729 | The time taken by a particle performing S.H.M. to pass from point ( boldsymbol{A} ) to ( boldsymbol{B} ) where its velocities are same is 2 seconds. After another 2 seconds it returns to ( mathrm{B} ) The time period of oscillation is (in seconds): A .2 B. 4 ( c cdot 6 ) ( D ) | 11 |
| 730 | A ( 1 mathrm{kg} ) mass executes SHM with an amplitude ( 10 mathrm{cm}, ) it takes ( 2 pi ) seconds to go from one end to the other end. The magnitude of the force acting on it at any end is : A . ( 0.1 mathrm{N} ) B. 0.2 N c. ( 0.5 mathrm{N} ) D. 0.05 N | 11 |
| 731 | A body is executing simple harmonic motion. At a displacement ( x ), its potential energy is ( E_{1} ) and at displacement ( y, ) its potential energy is ( boldsymbol{E}_{2} ) The potential energy ( mathrm{E} ) at a displacement ( (x+y) ) is A ( cdot E_{1}+E_{2} ) B. ( sqrt{E_{1}^{2}+E_{2}^{2}} ) c. ( E_{1}+E_{2}+2 sqrt{E_{1} E_{2}} ) D. ( sqrt{E_{1} E_{2}} ) | 11 |
| 732 | Assertion Graph between potential energy of a spring versus the extension or compression of the spring is a straight line Reason Potential energy of a stretched or compressed spring is proportional to square of extension or compression A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 11 |
| 733 | Two masses ( m_{1} ) and ( m_{2} ) are suspended together by a massless spring of spring constant ( k ) as shown in the figure. When the masses are in equilibrium, ( m_{1} ) is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of ( boldsymbol{m}_{2} ) A ( cdot sqrt{frac{k}{3 m_{2}}} ) B. ( sqrt{frac{k}{m_{2}}} ) c. ( sqrt{frac{k}{2 m_{2}}} ) D. ( sqrt{frac{k}{4} m_{2}} ) | 11 |
| 734 | A particle performs linear S.H.M. At a particular instant, velocity of the particle is ( ^{prime} u^{prime} ) and acceleration is ( ^{prime} alpha^{prime} ) while at another instant velocity is ( ^{prime} v^{prime} ) and acceleration is ( ^{prime} boldsymbol{beta}^{prime}(mathbf{0}<boldsymbol{alpha}<boldsymbol{beta}) . ) The distance between the two positions is A ( cdot frac{u^{2}-v^{2}}{alpha+beta} ) в. ( frac{u^{2}+v^{2}}{alpha+beta} ) c. ( frac{u^{2}-v^{2}}{alpha-beta} ) D. ( frac{u^{2}+v^{2}}{alpha-beta} ) | 11 |
| 735 | A body is executing SHM. If the force acting on the body is ( 6 mathrm{N} ) when the displacement is ( 2 mathrm{cm}, ) then the force acting on the body when the displacement is ( 3 mathrm{cm} ) in newton is: A. 6 N B. 9 N ( c .4 mathrm{N} ) D. ( sqrt{6} ) N | 11 |
| 736 | The displacement of a particle of mass ( 3 g m ) executing simple harmonic motion is given by ( y=3 sin (0.2 t) ) in ( s ) units. The kinetic energy of the particle at a point which is at a distance equal to ( frac{1}{3} ) of its amplitude from its mean position is A ( cdot 12 times 10^{-3} J ) ( J ) B . ( 25 times 10^{-3} J ) c. ( 0.48 times 10^{-3} J ) D. ( 0.24 times 10^{-3} mathrm{J} ) | 11 |
| 737 | The displacement of a particle in SHM is indicted by equation ( boldsymbol{y}= ) ( 10 sin left(20 t+frac{pi}{3}right) ) where ( y ) is in metres. The value of the maximum velocity of the particle will be? A. ( 100 mathrm{m} / mathrm{s} ) в. ( 150 mathrm{m} / mathrm{s} ) c. ( 200 m / s ) D. ( 400 mathrm{m} / mathrm{s} ) | 11 |
| 738 | For a particle performing SHM: A. The kinetic energy is never equal to the potential energy B. The kinetic energy is always equal to the potential energy C. The average kinetic energy in one time period is equal to the average potential energy in this period. D. The average kinetic energy in any time interval is never equal to average potential energy in that interval. | 11 |
| 739 | If at any instant of time the displacement of a harmonic oscillator is ( 0.02 mathrm{m} ) and its acceleration is ( 2 mathrm{ms}^{-2} ) its angular frequency at that instant will be A. ( 0.1 mathrm{rads}^{-1} ) B. ( 1 mathrm{rads}^{-1} ) ( mathbf{c} cdot 10 mathrm{rads}^{-1} ) D. ( 100 mathrm{rads}^{-1} ) | 11 |
| 740 | Vibrations, whose amplitudes of oscillations decrease with time are called ( A ). Free vibrations B. Forced vibrations c. Damped vibrations D. None of these | 11 |
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