Important Physics - Oscillations Questions for CBSE Class 11th

A particle is moving in a circle with uniform speed. Its motion is

A small mass executes linear SHM about O with amplitude a and period T. Its displacement from O at time T/8 after passing through O is

A vertical spring carries a 5 kg body and is hanging in equilibrium, an additional force is applied so that the spring is further stretched. When released from this position, it performs 50 complete oscillations in 25 s, with an amplitude of 5 cm. The additional force applied is

Resonance is an example

Which of the following figure represents damped harmonic motion

The acceleration due to gravity on the surface of the moon is 1 . 7 m / s 2 . What is the time period of a simple pendulum on the surface of moon if it’s time period on the surface earth is 3.5 seconds?(g on the surface of earth is 9 . 8 m / s 2 ).

A spherical solid sphere of mass m is tied to a light inextersible string. It is then whrited in a horizontal circular plane with constant angular speed then motion of the solid sphere is

Identify the correct statements for an SHM (i) Displacement leads velocity by 0.5 π (ii) Displacement lags velocity by π (iii) Acceleration leads velocity by 0.5 π (iv) Acceleration lags velocity by 0.5 π

Identify the correct statements for an SHM with amplitude A and maximum velocity = V (i)At position x = A 2 , velocity lags displacement by π 2 (ii)At position x = A 2 , velocity leads displacement by π 4 (iii)When velocity = V 2 , acceleration leads velocity by π (iv)When velocity = − V 2 , acceleration leads velocity by π 2

v – t graph of a graph of a particle in SHM is as shown in figure. Choose the correct

A particle is executing SHM in a straight line. Let x, v, and a represent instantaneous displacement, velocity and acceleration respectively. Graph of a 2 vs v 2 is

The angular frequency of oscillation for a particle executing angular oscillations with couple per unit twist of c is

The motion of a particle varies with time according to the relation y = 2 sin ω t + 2 cos ω t , then

A particle is executing simple harmonic motion. The equation of motion is given by d 2 x d t 2 + 4 x = 0 . Then their time period of oscillation is (in seconds)

A body of mass 5 g is executing SHM about a fixed point, with amplitude of 10 cm and maximum velocity 100 cm s –1 . Its velocity will be 50 cm s –1 at a distance of

A body in SHM has a maximum acceleration of 24 m s –2 and maximum velocity 16 m s –1 . The amplitude of SHM is nearly

Figure shows the variation of force acting on a particle of mass 400 g executing simple harmonic motion. The frequency of oscillation of the particle is

One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance x 0 from the block. The spring is then compressed by 2x 0 and released. The time taken to strike the wall is

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 (g=10 m/s 2 )

A particle is moving along the x-axis under the influence of a force given by F= – 5x + 15. At time t = 0, the particle is located at x = 6 and is having zero velocity. It takes 0.5 seconds to reach the origin for the first time. The equation of motion of the particle can be represented by

A simple pendulum is taken from the equator to the pole. Its period

Two masses m 1 and m 2 are suspended together by a massless spring of constant k. When the masses are in equilibrium, m 1 is removed without disturbing the system. Then the angular frequency of oscillation of m 2 is

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m then the time period becomes 5 4 T . The ratio of m M is

A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance x. Now the combined mass will oscillate on the spring with period

A spring having a spring constant ‘K’ is loaded with a mass ‘m’. The spring is cut into two equal parts and one of these is loaded again with the same mass. The new spring constant is

The length of a spring is l and its force constant is k. When a weight W is suspended from it, its length increases by x. If the spring is cut into two equal parts and put in parallel and the same weight W is suspended from them, then the extension will be

A mass m =100 gms is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equal to 0.16 metre and time period equal to 2 sec. Initially the mass is released from rest at t = 0 and displacement x = − 0 .16 metre. The expression for the displacement of the mass at any time t is

A block is placed on a frictionless horizontal table. The mass of the block is m and springs are attached on either side with force constants K 1 and K 2 . If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will be

A mass m attached to a spring oscillates every 2 sec. If the mass is increased by 2 kg, then time-period increases by 1 sec. The initial mass is

Two springs of constant k 1 and k 2 are joined in series. The effective spring constant of the combination is given by

To make the frequency double of a spring oscillator, we have to

If a body of mass 0.98 kg is made to oscillate on a spring of force constant 4.84 N/m, the angular frequency of the body is

A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force Fsinωt . If the amplitude of the particle is maximum for ω = ω 1 and the energy of the particle is maximum for ω = ω 2 , then (where ω 0 natural frequency of oscillation of particle)

Amplitude of a wave is represented by A = c a + b − c .Then resonance will occur when

A simple pendulum is set into vibrations. The bob of the pendulum comes to rest after some time due to

A simple pendulum oscillates in air with time period T and amplitude A. As the time passes

A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross-section and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to

On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant K, the period of oscillation of the body (assuming the springs as massless) is

In SHM there is always a constant ratio between displacement of the body and it’s

When a block is suspended by means of spring the restoring force per unit displacement of the block is called

For a simple harmonic oscillator the frequency of oscillation is independent of

A particle moves on the axis according to the equation, x = x o S i n 2 ω t . The motion is simple harmonic

A man measure the period of a simple pendulum inside a stationary lift and finds it to be T seconds. If the lift accelerates upwards with an accelerating g/4, then the period of the pendulum will be

If the maximum acceleration of a SHM is ‘a’ and the maximum velocity is ‘b’ , then amplitude of vibration is

The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2×10 -2 cos(πt)m . The time at which the maximum speed first occurs is

Springs of spring constants K,2K,4K,8K,…,many are connected in series. A mass ‘m’ is attached to one end. The system is allowed to oscillation. The time period is approximately.

A freely falling body takes 2 seconds to reach the ground on a planet, when it is dropped from a height of 8m. If the period of a simple pendulum is π seconds on the planet. The length of a pendulum is ( in centi meter)

At what displacement is the kinetic energy of a particle performing simple harmonic motion of amplitude 10 cm is three times it’s potential energy?

A body executing SHM has a maximum velocity 1 m/sec, maximum acceleration of 4 m/ s e c 2 . It’s amplitude in meters is

Find the limiting values of amplitude of a force damped oscillator, when (i) ω d < < ω 0 and (ii) ω d > > ω 0 .

In the following examples, which represents periodic but not simple harmonic motion a) Rotation of earth about its own axis b) Revolution of moon around the earth c) Swinging pendulum from a fixed axis d) Marble released from top of a smooth hemispherical bowl

Graph of v 2 V s x 2 for a particle executing SHM is

A physical pendulum pivoted at a point executes angular oscillations. Its mass m,has its centre of mass at distance r from the point of suspension. If its moment of inertia is I, then its angular frequency is

A cylindrical plastic bottle of negligible mass is filled with 500ml of water. On slightly pressing it downwards and releasing, it starts executing simple harmonic motion. If radius of the bottle is 5cm, then its angular frequency of oscillation is close to ( d e n s i t y o f w a t e r ρ = 10 3 k g / m 3 )

A body executes S.H.M. under the action of a force F 1 with a time period 7/6 seconds. If the force is changed to F 2 it executes S.H.M. with time period 7/8 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 body is dropped on to the floor from a height h. It makes elastic collision with the floor bouncing back to the same height. The frequency of oscillation of its periodic motion is

A particle under simple harmonic motion has a time period of T. The time period with which the potential energy changes is

A particle of mass m is moving in a potential well, for which the potential energy is given by U ( x ) = U 0 1 − sin b x where U 0 and a are constants. Then for small oscillations

An sphere of moment of inertia 4 kg/m 2 passing through the diameter. It is suspended from a steel wire of negligible mass, whose couple per unit twist is 81 Nm/rad. It is allowed to execute small angular oscillations. The angular frequency of oscillation is

Two particles executes S.H.M. of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions, and each time their displacement is half of their amplitude. The phase difference between them is

Two particles move parallel to the x-axis about the origin with same amplitude ‘a’ and frequency ω . 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?

Two simple pendulum first of bob mass M 1 and length L 1 second of bob mass M 2 and length L 2 . M 1 = M 2 and L 1 =2L 2 .If these vibrational energy of both is same. Then which is correct

A uniform semicircular ring having mass m and radius r is hanging at one of its ends freely as shown in figure. The ring is slightly disturbed so that it oscillates in its own plane. The time period of oscillation of the ring is

Two particles A and B execute simple harmonic motion according to the equations ) y 1 = 3 sin ⁡ ω t and y 2 = 4 sin ⁡ ω t + π 2 + 3 sin ⁡ ω t The phase difference between them is

A particle of mass m is executing oscillations about the origin on the X-axis with amplitude A. Its potential energy is given as U(x)= β x 4 , where β is a positive constant. The x-coordinate of the particle, where the potential energy is one-third of the kinetic energy, is

An air chamber of volume V, has a long of cross-sectional area A. A ball of mass m is fitted smoothly in the neck. The bulk modulus of air is B. If the ball is pressed down slightly and released, the time period of its oscillation is

A particle executes SHM along x-axis with a frequency 25 π Hz . The mass of the particle is 0.20 kg. At the position x = 0.04 m, the kinetic energy is 0.50 J and the potential energy is 0.40 J. The potential energy is zero at mean position. The amplitude of oscillations (in cm) is .

A particle in S.H.M. is described by the displacement function x ( t ) = acos ( ωt + θ ) . If the initial ( t = 0 ) position of the particle is 1 cm and its initial velocity is π cm / s . The angular frequency of the particle is π rad / s , then it’s amplitude is

A system exhibiting S.H.M. must possess

A particle is moving in a circle with uniform speed. Its motion is

A simple pendulum performs simple harmonic motion about X = 0 with an amplitude A and time period T. The speed of the pendulum at X = A 2 will be

A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/sec. Its velocity will be 50 cm/sec at a distance

A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The magnitude of the velocity in msec − 1 at the centre of oscillation is

A S.H.M. has amplitude ‘a’ and time period T. The maximum velocity will be

A particle is executing the motion x = Acos ( ω t − θ ) . The maximum velocity of the particle is

Displacement between maximum potential energy position and maximum kinetic energy position for a particle executing S.H.M. is

Consider the following statements. The total energy of a particle executing simple harmonic motion depends on its (1) Amplitude (2) Period (3) Displacement Of these statements

The amplitude of a particle executing SHM is made three-fourth keeping its time period constant. Its total energy will be

A body is executing Simple Harmonic Motion. At a displacement x its potential energy is E 1 and at a displacement y its potential energy is E 2 . The potential energy E at displacement ( x + y ) is

What is constant in S.H.M.

To make the frequency double of an oscillator, we have to

The equation of a simple harmonic motion is X = 0 .34 cos ( 3000 t + 0 .74 ) where X and t are in mm and sec. The frequency of motion is

In a simple pendulum, the period of oscillation T is related to length of the pendulum l as

The time period of a simple pendulum when it is made to oscillate on the surface of moon

The height of a swing changes during its motion from 0.1 m to 2.5 m. The minimum velocity of a boy who swings in this swing is

A simple pendulum hanging from the ceiling of a stationary lift has a time period T 1 . When the lift moves downward with constant velocity, the time period is T 2 , then

A simple pendulum is vibrating in an evacuated chamber, it will oscillate with

What effect occurs on the frequency of a pendulum if it is taken from the earth surface to deep into a mine

Two identical spring of constant K are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be

The vertical extension in a light spring by a weight of 1 kg suspended from the wire is 9.8 cm. The period of oscillation

If a watch with a wound spring is taken on to the moon, it

A mass m performs oscillations of period T when hanged by spring of force constant K. If spring is cut in two parts and arranged in parallel and same mass is oscillated by them, then the new time period will be

Resonance is an example of

One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance x 0 from the black. The spring is then compressed by 2 x 0 and released. The time taken to strike the wall is

Which of the following function represents a simple harmonic oscillation

The amplitude of a damped oscillator becomes half in one minute. The amplitude after 3 minute will be 1 X times the original, where X is

A particle is executing S.H.M. Then the graph of acceleration as a function of displacement is

A particle of mass m oscillates with simple harmonic motion between points x 1 and x 2 , the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graph

The velocity-time diagram of a harmonic oscillator is shown in the adjoining figure. The frequency of oscillation is

Graph between velocity and displacement of a particle, executing S.H.M. is

For a simple pendulum the graph between L and T will be.

In case of a simple pendulum, time period versus length is depicted by

One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and the Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate. Find the time period of oscillation.

The motion of a particle in SHM is of

A girl swings on a cradle in sitting position, if she stands, the time period of cradle

Which of the following four statements is false?

Calculate the change in the length of a simple pendulum of length 1m, when its period of oscillation changes from 2 sec to 1.5 sec

What happens to the energy of a simple harmonic oscillator if its amplitude is doubled?

The motion of a torsional pendulum is

Two springs of force constant 1000 N/m & 2000 N/m are stretched by same force. The ratio of their respective potential energies is

When a mass of 0.5 kg is suspended from the free end of spring it stretches the spring by 0.2 m. This mass is removed and 0.25 kg mass is attached to the same free end of the spring. If the mass is pulled down and released. What is it’s time period? ( g = 10 m s – 1 )

A particle performs simple harmonic motion with amplitude A. It’s speed is trebled at the instant that it is at a distance 2A/3 from equilibrium position. The new amplitude of the motion is

A simple pendulum has time period ‘ T 1 ‘. The point of suspension is now moved upwards according to the relation y = K T 2 . ( K = 1 m / s e c 2 ) where y is the vertical displacement. The time period now becomes ‘ T 2 ‘. Then find the ratio of T 1 2 T 2 2 ( g = 10 m / s e c 2 )

A simple pendulum has a time period ‘T’ in vacuum. It’s time of period when it is completely immersed in a liquid of density one-eighth of the density of the material of the bob is

The total mechanical energy of a harmonic oscillator of amplitude 1 m and force constant 200 N/m is 150 J then

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then its periodic time (in seconds) is

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by ‘m’ the time period becomes 5 T 3 , then the ratio of m M is

The time period of simple pendulum is ‘T’ when the length increases by 10cm, its period is T 1 . When the length is decreased by 10 cm, its time period is T 2 . When the relation between T, T 1 and T 2 is

Springs of force constants K, 2K, 4K, 8K, 16K,… are connected in series. The mass ‘m’ kg is attached to the lower end of the last spring and the system is allowed to vibrate. The frequency of oscillation is

A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them is

A body executing SHM has a maximum acceleration equal to 48 m / s e c 2 and maximum velocity equal to 12 m/sec. The amplitude of SHM is

A man measure the period of a simple pendulum inside a stationary lift and finds it to be T seconds. If the lift accelerates upwards with an accelerating g/4, then the period of the pendulum will be

Two particles are executing simple harmonic motion at an instant of time ‘t’ their displacements are y 1 =acos(wt) and y 2 =asin(wt) . Then the phase difference between y 1 and y 2 is

The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2×10 -2 cos(πt) meter . The time at which the maximum speed first occurs is

The displacement y of a wave travelling in the x direction is given by y = 10 – 4 sin ( 600 t – 2 x + π / 3 ) metre. When x is expressed in metres and t in seconds. The speed of the wave motion in m s – 1 is

A body executes simple harmonic motion the potential energy (PE). The kinetic energy (KE) and total energy (TE) are measured as a function of displacement x. Which of the following statement is true?

A particle performs simple harmonic motion with amplitude A. It’s speed is trebled at the instant that it is a distance 2A/3 from equilibrium position. The new amplitude of the motion is

The displacement of two identical particles executing S.H.M are represented by equation x 1 = 4 sin ( 10 t + π / 6 ) a n d x 2 = 5 cos ( w t ) . For what value of ‘w’ energy of both the particles is same.

Two particles A and B of equal masses are suspended from two mass less spring of spring constants K 1 and K 2 respectively. If the maximum velocities, during oscillations are equal. The ratio of amplitudes of ‘A’ and ‘B’.

Springs of spring constants K,2K,4K,8K,……. are connected in series. A mass ‘m’ is attached to one end. The system is allowed to oscillation. The time period is approximately.

A simple pendulum has a time period ‘T’ in vacuum. It’s time of period when it is completely immersed in a liquid of density one-eighth of the density of the material of the bob is

The velocity of particle undergoing SHM at the mean position is 4m/s. Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude

A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th the radius of the earth. What will be its time period oscillation?

A body executing SHM has a maximum velocity 1 m/sec, maximum acceleration of 4 m / s e c 2 . It’s amplitude in meters is

A body oscillates with S.H.M according to the equation x=5cos 2πt+ π 4 metre its instantaneous displacement at t=1 sec is

A body of mass 40 g executes simple harmonic motion of amplitude 2 cm. If the time period is 0.20 seconds. What would be the total mechanical energy of system?

The displacement y of a particle in a medium can be expressed as y=10 -6 sin 100t+20x+ π 4 m where t is in second and x in metre. The speed of the wave is

The acceleration due to gravity on moon is 1 6 th of that on the earth. If the length of the seconds pendulum is 60 cm on the earth, what is its length on the moon?

A simple pendulum has time period 3 seconds. The pendulum is completely immersed in a non viscous liquid whose density is 1 10 th of that of the material of the bob. Then time period of that pendulum is

A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period T 1 and, (ii) back and forth in a direction perpendicular to its plane, with a period T 2 . The ratio T 1 T 2 will be

In a spring- mass system with mass m = 0.5kg and k = 200 Nm –1 , what is the value of damping coefficient for critical damping?

A 21.2kg object oscillates at the end of a vertical spring that has a spring constant 20500 Nm –1 . The effect of air resistance is represented by the damping coefficient b = 2 Nsm –1 . Find the time interval that elapses while the energy of the system drops to 10% of its initial value (Given, ln 10 = 2.302)

A particle of mass m moves in one dimension, whose potential energy varies as U x = − a x 2 + b x 4 , where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to

A pendulum of length 1cm hangs from an inclined wall. Suppose that this pendulum is released at an initial angle of 10 ∘ and it bounces off the wall elastically when it reaches an angle of − 5 ∘ as shown in the figure. (Take g = π 2 m / s 2 ) The period of this pendulum is (in second)

A uniform square plate of mass m is supported in a horizontal plane by a vertical pin at B and is attached at A to a spring of constant K. If corner A is given a small displacement and released, determine the period of the resulting motion

A body of mass m is hanging from a wire of length l. It is pulled by a distance Δ l and left. It executes simple harmonic motion. If the young’s modulus of the wire is Y, area of cross-section is A, then the frequency of oscillation is

Potential energy of a particle executing SHM is U at an instant. Magnitude of acceleration is a at the same instant. If x is the magnitude of displacement of the particle

For a particle executing simple harmonic motion, the kinetic energy K is given by, K 0 cos 2 ⁡ ω t .The maximum value of PE is

The equation of motion of a particle executing simple harmonic motion is given by a + 36 π 2 x = 0 , where a is acceleration and x is displacement. Then, its frequency of oscillation is (in hertz)

The period of a particle in SHM is 8 sec. At t = 0, it is at the mean position. The ratio of the distances travelled by it in the first one second and in the 2 nd one second is

In an SHM, during the journey of the particle from mean position to extreme position, what is the average acceleration on the particle (Amplitude = A, angular frequency = ω )?

A particle executing simple harmonic motion is given by y = 8 sin π 10 t − 1 2 where y is in metre, and t is in seconds. Then the average speed of the particle during one full oscillation is

Displacement-time equation of a particle executing SHM is x = 10 sin π 3 t + π 6 c m . The distance covered by particle in first 3 seconds is

The speed v of a particle moving along a straight line when it is at a distance x from a fixed point on the line is given by v 2 = 144 − 9 x 2 . Select the correct alternatives

PE of a particle is U ( x ) = a x 2 − b x . Then the time period of small oscillation.

The mean kinetic energy of a particle executing SHM over one full oscillation, given mass of particle = m, amplitude = A, angular frequency = ω ).

A hanging object cannot be considered as a point mass and it oscillates about a fixed axis which does not pass through its centre of mass uniform rod of mass M and length L. The rod is pivoted at one end and hangs vertically in equilibrium with its centre of mass vertically below the point of suspension. The rod is slightly at the lower end and released. It then oscillates in a vertical plane in a simple harmonic manner, at any instant angular displacement of the rod from its vertical position is θ assuming small θ , time period of oscillation of the rod is (Here I is the moment of inertia of the rod about the axis about which the rod oscillates )

A body is oscillating simple harmonically with a period T = 2 s. How much time would it take for its kinetic energy to decrease from K max to 75 % of K max ?

A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is

The metallic bob of a simple pendulum has the relative density ρ . The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by

A simple pendulum has time period T 1 The point of suspension is now moved upward according to equation y = k t 2 , where k = 1 m/sec 2 . If new time period is T 2 then ratio T 1 2 T 2 2 will be

For a particle executing S.H.M. the displacement x is given by x = A cos ⁡ ω t . identify the graph which represents the variation of potential energy (P.E.) as a function of time t and displacement x

Figure shows the variation of force acting on a particle of mass 400 g executing simple harmonic motion. The frequency of oscillation of the particle is

An object of mass 0.2 kg executes simple harmonic along X-axis with frequency of 25 π Hz . At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation in meter is equal to

A particle is executing SHM according to the equation x = A cos ⁡ ω t . Average speed of the particle during the interval 0 ≤ t ≤ π 6 ω .

A metre stick swinging in vertical plane about a fixed horizontal axis passing through its one end undergoes small oscillation of frequency f 0 . If the bottom half of the stick were cut off, then its new frequency of small oscillation would become

A system of two identical rods (L-shaped) of m ass m and length I are resting on a peg P as shown in the figure. If the system is displaced in its plane by a small angle, find the period of oscillations.

A charged particle is deflected by two mutually perpendicular oscillating electric fields such that the displacement of the particle due to each one of them is given by x = A sin ⁡ ( ω t ) and y = A sin ⁡ ω t + π 6 respectively. The trajectory followed by the charged particle is

A particle is executing a motion in which its displacement as a function of time is given by x = 3 sin ⁡ ( 5 π t + π / 3 ) + cos ⁡ ( 5 π t + π / 3 ) where x is in m and t is in s. Then the motion is

The string of a simple pendulum is replaced by a uniform rod of length L and mass M while the bob has a mass m. It is allowed to make small oscillations. Its time period is

A cork floating on the pond water executes a simple harmonic motion, moving up and down over a range of 4 cm. The time period of the motion is 1 s. At t=0, the cork is at its lowest position of oscillation, the position and velocity of the cork at t = 10.5 s, would be

A solid right circular cylinder of weight 10 kg and cross- section al area 100 cm 2 is suspended by a spring, where k=11 kg/cm, and hangs partially submerged in water of density 1000 kg/m 3 as shown in figure. What is its period when it makes simple harmonic vertical oscillations? (Take g=10 m/s 2 )

A particle starts its SHM on a line at initial phase of π / 6 .It reaches again the point of start after time t. lt crosses yet another point P on the same line at successive intervals 2t and 3t respectively. Find the amplitude of the motion, if the particle crosses the point of start at speed 2 m/s:

From the variation of potential energy in the direction of small oscillation of a simple pendulum, find the effective spring constant for the simple pendulum, where m is mass of the bob, l is length of the simple pendulum.

A body performs SHM along the straight line segment ABCDE with c as the mid point of segment AE (A and E are the extreme position for the SHM). Its kinetic energies at B and D are each one fourth of its maximum value. The length of segment AE is 2R, if the distance between B and, D is found to be x R . The value of x is .

A constant force produces maximum velocity v on the block connected to the spring of force constant k as shown in the figure. When the force constant of spring becomes 4k,the maximum velocity of the block is (block is at rest when spring is relaxed):

Three identical ideal springs, each of force constant k are joined to three identical balls (each of mass m) as shown in the figure. O is centroid of the triangle. Initially, each of the springs is in its natural length. Now all the three balls are simultaneously given small displacements of equal magnitude along the directions shown in the figure. The oscillation frequency of balls is 1 2 π β k 2 m . The value of β is .

A force F = − 4 x + 8 (inN) is acting on block where x is the position of block in metres. The energy of oscillation is 32 J. The block oscillates between two points, out of which value of position of one point (in m) is an integer from 0 to 9. The value of x is .

A simple pendulum of length 1 m is allowed to oscillate with amplitude 2 o .It collides elastically with a wall inclined at 1 o to the vertical. Its time period will be: (use g = π 2 )

A physical pendulum is positioned so that its centre of gravity is above the suspension point. when the pendulum is released it passes the point of stable equilibrium with an angular velocity ω . The period of small oscillations of the pendulum is

A particle executing simple harmonic motion along y-axis has its motion described by the equation y = Asin ( ω t ) + B . The amplitude of the simple harmonic motion is

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

The periodic time of a body executing simple harmonic motion is 3 sec. After how much interval from time t = 0, its displacement will be half of its amplitude

A particle is executing S.H.M. If its amplitude is 2 m and periodic time 2 seconds, then the maximum velocity of the particle will be

A body is executing S.H.M. When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocity of the body is 10 cm/sec and 8 cm/sec. Then the time period of the body is

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 . Then angular velocity will be

If a particle under S.H.M. has time period 0.1 sec and amplitude 2 × 10 – 3 m . It has maximum velocity

A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceleration at a distance of 2 cm from the mean position is 8 cm / s 2 . The maximum speed of the particle is

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is

The angular velocities of three bodies in simple harmonic motion are ω 1 , ω 2 , ω 3 with their respective amplitudes as A 1 , A 2 , A 3 . If all the three bodies have same mass and velocity, then

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes 8 3 cm / s , will be

The displacement equation of a particle is x = 3 sin 2 t + 4 cos 2 t . The amplitude and maximum velocity will be respectively

Velocity at mean position of a particle executing S.H.M. is v, they velocity of the particle at a distance equal to half of the amplitude

A particle moves such that its acceleration a is given by a = − bx , where x is the displacement from equilibrium position and b is a constant. The period of oscillation is

The equation of motion of a particle is d 2 y dt 2 + Ky = 0 , where K is positive constant. The time period of the motion is given by

The time period of a simple pendulum is 2 sec. If its length is increased 4 times, then its period becomes

The time period of a simple pendulum in a lift descending with constant acceleration g is

The acceleration due to gravity at a place is π 2 m / sec 2 . Then the time period of a simple pendulum of length one metre is

The motion of a particle executing S.H.M. is given by x = 0 .01 sin 100 π ( t + . 05 ) , where x is in metres and time is in seconds. The time period is

In a seconds pendulum, mass of bob is 30 gm. If it is replaced by 90 gm mass. Then its time period will

Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is 7 : 8, then the ratio of lengths of the two pendulums will be

A simple pendulum is attached to the roof of a lift. If time period of oscillation, when the lift is stationary is T. Then frequency of oscillation, when the lift falls freely, will be

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g 3 is

The amplitude of an oscillating simple pendulum is 10cm and its period is 4 sec. Its speed after 1 sec after it passes its equilibrium position, is

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of 10cm (g = 9.8 m/s2)

There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes g / 4 , then the new time period of the pendulum is

The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g / 3 , then the time period of the pendulum is

A simple pendulum hangs from the ceiling of a car. If the car accelerates with a uniform acceleration, the frequency of the simple pendulum will

The periodic time of a simple pendulum of length 1 m and amplitude 2 cm is 5 seconds. If the amplitude is made 4 cm, its periodic time in seconds will be

A mass m is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are k 1 and k 2 respectively. When set into vertical vibrations, the period will be

In the figure, S 1 and S 2 are identical springs. The oscillation frequency of the mass m is f . If one spring is removed, the frequency will become

A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system if a mass 4 m is suspended from the same spring

The frequency of oscillation of the springs shown in the figure will be

Two springs of force constants K and 2K are connected to a mass as shown below. The frequency of oscillation of the mass is

A particle at the end of a spring executes simple harmonic motion with a period t 1 , while the corresponding period for another spring is t 2 . If the period of oscillation with the two springs in series is T, then

Infinite springs with force constant k, 2k, 4k and 8k…. respectively are connected in series. The effective force constant of the spring will be

When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2.0 kg block to the spring and if the block is pulled through 10 cm and released the maximum velocity in it in m/s is : (Acceleration due to gravity = 10 m / s 2 )

Two springs with spring constants K 1 = 1500 N / m and K 2 = 3000 N / m are stretched by the same force. The ratio of potential energy stored in spring will be

The time period of a mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

A spring executes SHM with mass of 10kg attached to it. The force constant of spring is 10N/m.If at any instant its velocity is 40cm/sec, the displacement will be (where amplitude is 0.5m)

The S.H.M. of a particle is given by the equation y = 3 sinω t + 4 cosω t . The amplitude is

If the displacement equation of a particle be represented by y = AsinPT + BcosPT , the particle executes

The resultant of two rectangular simple harmonic motions of the same frequency and unequal amplitudes but differing in phase by π 2 is

Two mutually perpendicular simple harmonic vibrations have same amplitude, frequency and phase. When they superimpose, the resultant form of vibration will be

The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of π results in the displacement of the particle along

A S.H.M. is represented by x = 5 2 ( sin 2 πt + cos 2 πt ) . The amplitude of the S.H.M. is

The displacement of a particle varies with time as x = 12 sinωt − 16 sin 3 ωt (in cm). If its motion is S.H.M., then its maximum acceleration is

The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t 0 in air. Neglecting frictional force of water and given that the density of the bob is (4/3) ×1000 kg/m3. What relationship between t and t 0 is true [AIEEE 2004]

A simple pendulum has time period T 1 . The point of suspension is now moved upward according to equation y = kt 2 where k = 1 m / sec 2 . If new time period is T 2 then ratio T 1 2 T 2 2 will be

A disc of radius R and mass M is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be

Three masses 700g, 500g, and 400g are suspended at the end of a spring a shown and are in equilibrium. When the 700g mass is removed, the system oscillates with a period of 3 seconds, when the 500 gm mass is also removed, it will oscillate with a period of

Two simple pendulums whose lengths are 100 cm and 121 cm are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again

The displacement time graph of a particle executing S.H.M. is as shown in the figure The corresponding force-time graph of the particle is

As a body performs S.H.M., its potential energy U varies with time as indicated in

Acceleration A and time period T of a body in S.H.M. is given by a curve shown below. Then corresponding graph, between kinetic energy (K.E.) and time t is correctly represented by

Which of the following displacement-time graphs represent damped harmonic oscillation?

Total energy of particle performing SHM depends on

An oscillating simple pendulum is arranged in a vacuum jar, then

If iron sphere is replaced by wooden sphere of same mass, time period of simple pendulum

A simple pendulum is set into vibrations. The bob of the pendulum comes to rest after some time due to

If a pendulum clock is shifted from earth to the surface of moon. Then it

The frequency of oscillation of a simple pendulum suspended in a satellite that revolve around the earth is

A simple pendulum has a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off. The period of the pendulum will

The phase of a simple harmonic motion at t=0 is called

The phase difference between velocity and acceleration of simple harmonic oscillator

The displacement of a particle in SHM is in opposite phase with

The motion of a torsional pendulum is

At the equilibrium position of a particle executing SHM

On an average a human heart is found to beat 75 times in a minute. Calculate its frequency

A mass of 2kg attached to a spring of force constant 260 N m – 1 makes 100 oscillations. What is the time taken for 100 oscillations?

A simple pendulum hanging freely and at rest is vertical because in that position

The phase difference between acceleration and displacement of S.H.M.

Which of the following functions represents a SHM?

The amplitude of vibration of a particle is given by a m = a o a ω 2 − b ω + c , where a o , a, b & c are positive. The condition for a single resonant frequency.

A particle is subjected to two SHM’s x 1 = A 1 S i n ω t a n d x 2 = A 2 S i n ( ω t + π / 4 ) . The resultant SHM will have an amplitude of

Two simple harmonic oscillators with amplitudes in the ratio 1:2 are having the same total energy, the ratio of their frequencies is

The periods of a pendulum on two planets are in the ratio 3:4, the acceleration due to gravity on them are in the ratio

The time period of stiffer spring as compared to that of a soft spring is

Two particles are executing simple harmonic motion at an instant of time ‘t’ their displacements are y 1 =acos(wt) and y 2 =asin(wt) . Then the phase difference between y 1 and y 2 is

The total energy of a vibrating particle in S.H.M is E. If it’s amplitude and time period are doubled, it’s total energy will be

The displacement y of a wave travelling in the x direction is given by y = 10 – 4 sin ( 600 t – 2 x + π / 3 ) metre. When x is expressed in metre and t in seconds. The speed of the wave motion in m s – 1 is

A body executes simple harmonic motion the potential energy (PE), the kinetic energy (KE) and total energy (TE) are measured as a function of displacement x. Which of the following statement is true?

A simple pendulum of length ‘ L 1 ‘ has a time period ‘ T 1 ‘ and another pendulum of length ‘ L 2 ‘ has time period ‘ T 2 ‘. Then the time period of pendulum of length ( L 1 + L 2 ) is

If a spring of force constant ‘K’ is cut into three equal parts then the force constant of each part will be

Restoring force on the bob of a simple pendulum of mass 100 g when its amplitude is 1° is (here acceleration due to gravity =10m s – 2 )

A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th The radius of the earth is R. What will be its time period of oscillation?

The velocity of particle undergoing SHM at the mean position is 4m/s. Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude

A body executing SHM at a displacement ‘x’ it’s PE is E 1 ,at a displacement ‘y’ its PE is E 2 . The PE at a displacement (x+y) is

Two springs of force constant 2000 N/m and 4000 N/m are stretched by same force. The ratio of their respective potential energies is

Time period of a spring mass system is T. If this spring is cut into two parts whose lengths are in the ratio 1:3 and the same mass is attached to the longer part. The new time period will be

A body oscillates with S.H.M according to the equation x=5cos 2πt+ π 4 its instantaneous displacement at t=1 sec is

The time period of a simple pendulum inside a stationary lift is 5 sec. What will be the time period when the lift moves upward with an acceleration g/4?

A body of mass 40 grams executes simple harmonic motion of amplitude 2 cm. If the time period is 0.20 seconds. What would be the total mechanical energy of system?

The displacement of a particle is represented by the equation y=3cos π 4 -2wt . The motion of the particle is

In a simple harmonic oscillator, at the mean position is

When a particle is performing linear SHM its KE is two times its PE at a position ‘A’ and its PE is two times its KE at another position ‘B’. Find ratio of KE A to KE B

The acceleration due to gravity on moon is 1 6 th of that on the earth. If the length of the seconds pendulum is 96 cm on the earth, what is its length on the moon?

The period of a simple pendulum measured inside a stationary lift is ‘T’. If the lift starts moving upwards with a acceleration g/3. What will be the time period.

If a spring has time period ‘T’ and cut into ‘n’ equal parts then the time period of each part will be

Tangential acceleration on a bob of a simple pendulum of mass 100 grams when its amplitude is 1 o is (here g=10m/ s 2 )

On a planet a body is left fall freely from a height 2 metres reaches the ground 2 seconds if the length of a simple pendulum is 2 metres on that planet, its time period is

The mass and diameter of a planet are twice that of the earth. What will be the time period of oscillation of a pendulum on this planet, if it is a seconds pendulum on earth

A particle of mass m executes SHM with amplitude ‘a’ and frequency f, the average kinetic energy during motion from the position of equilibrium to the end is

Two springs of spring constants 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have potential energy in the ratio

The force on a body executing SHM is 2 N when the displacement is 2 cms if the amplitude of oscillations is 5 cm. What is the total energy associated with the SHM?

A particle execute SHM from the mean position. Its amplitude is A, its time period is ‘T’. At what displacement, its speed is half of its maximum speed.

The total energy of a vibrating particle in S.H.M is E. If it’s amplitude is tripled and time period are doubled, it’s total energy will be

A simple pendulum of length L 1 has a time period ‘ T 1 ‘ and another pendulum of length ‘ L 2 ‘ has time period ‘ T 2 ‘. Then the time period of pendulum of length ( L 1 – L 2 ) is

If a spring of force constant ‘K’ is cut into three equal parts then the force constant of each part will be

Restoring force on the bob of a simple pendulum of mass 100 g when its amplitude is 2° is(g=10m/ s – 2 )

A body executing SHM at a displacement ‘x’ it’s PE is E 1 ,at a displacement ‘y’ its PE is E 2 . The PE at a displacement (x+y) is

Two springs of force constant 3000 N/m and 6000 N/m are stretched by same force. The ratio of their respective potential energies is

At what displacement is the KE of a particle performing SHM of amplitude 10 cm equal to three times it’s PE?

The total mechanical energy of a harmonic oscillator of amplitude 1 m and force constant 200 N/m is 100 J then

Time period of a spring mass system is T. If this spring is cut into two parts whose lengths are in the ratio 1:3 and the same mass is attached to the longer part. The new time period will be

The time period of a simple pendulum inside a stationary lift is 5 sec. What will be the time period when the lift moves upward with an acceleration g/4?

The displacement of a particle is represented by the equation y=3cos π 4 -2ωt . The motion of the particle is

When a particle is performing linear SHM its KE is two times its PE at a position ‘A’ and its PE is two times its KE at another position ‘B’. Find ratio of KE A to KE B

In a simple harmonic oscillator, at the mean position is

The period of a simple pendulum measured inside a stationary lift is ‘T’. If the lift starts moving upwards with a acceleration g/3. What will be the time period.

The time period of simple pendulum is ‘T’ when the length increases by 20cm, its period is T 1 . When the length is decreased by 20 cm, its time period is T 2 . When the relation between T, T 1 and T 2 is

Springs of constants K, 2K, 4K, 8K, 16K,… are connected in series. The mass ‘m’ kg is attached to the lower end of the last spring and the system is allowed to vibrate. The frequency of oscillation is

If a spring has time period ‘T’ and cut into ‘n’ equal parts then the time period of each part will be

Tangential acceleration on a bob of a simple pendulum of mass 100 grams when its amplitude is 2 0 is

On a planet a body is left fall freely from a height 2 metres reaches the ground 2 seconds if the length of a simple pendulum is 2 metres on that planet, its time period is

The mass and diameter of a planet are 4 times that of the earth. What will be the time period of oscillation of a pendulum on this planet, if it is a seconds pendulum on earth

A particle of mass ‘ m ’ is fixed to one end of light spring having force constant ‘ k ’and unstretched length ‘ l ’. The other end is fixed. The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is

A spring mass system (mass ‘m’, spring constant ‘k’ and natural length l 0 ) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of disc. If 5g mass of the body attached to the spring having spring constant 20 N m − 1 . If the disc together with spring mass system, rotates about its axis with an angular velocity of 5 rad.s − 1 . k > > m ω 2 , then the percentage change in the length of the spring is

A rod of mass ‘M’ and length ‘2L’ is suspended at its middle by a wire. It exhibits torsional oscillations , If two masses each of mass ‘m’ are attached at distance L 2 from centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is closed to

A particle executes simple harmonic motion with an amplitude of 5cm. When the particle is at 4cm from the mean position. The magnitude of its velocity in SI units is equal to that of its acceleration. Then its periodic time in seconds is

A particle undergoing simple harmonic motion has time dependent displacement given by x t = A sin π t 90 . The ratio of kinetic to potential energy of this particle at time t = 30 s will be

A simple pendulum of length 1m is oscillating with an angular frequency 10 rad s -1 . The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad s -1 and an amplitude of 10 -2 m . The relative change in the angular frequency of the pendulum is best given by(take acceleration due to gravity g=10m/ s 2 )

A damped harmonic oscillator has a frequency of 5 oscillations per second.The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to (1/1000) of the original amplitude is close to

A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 10 16 t h of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is

The amplitude of vibration of a particle is given by a m = a 0 a ω 2 − b ω + c where a 0 , a , b and c are positive. The condition for a single resonant frequency is

A particle moves so that its acceleration a is given by a = − b x , where x is displacement from the equilibrium position and b is a constant. The period of oscillation is

A particle moves in a circular path with a uniform speed. Its motion is

The potential energy of a particle U x executing SHM is given by

Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point a distance of the ‘ h‘ directly above the tunnel, the motion of the particle is

A block of mass m is projected towards a spring with velocity . The force constant of the spring is K. The block is projected from a distance from the free end of the spring .The collision between block and the wall is completely elastic. Match the following columns Column – I Column – II A) maximum compression of the spring P) − K V 0 2 m B) Energy of oscillations of block Q) m V 0 2 k C) Time period of oscillations R) 1 2 m V 0 2 D) maximum acceleration of the S) 2 l V 0 + π m k

The displacement time graph of a particle executing S.H.M. is given in figure: (sketch is schematic and not to scale) Which of the following statements is/are true for this motion? (A) The force is zero at t = 3 T 4 (B) The acceleration is maximum at t = T (C) The speed is maximum at t = T 4 (D) The P.E. is equal to K.E. of the oscillation at t = T 2

A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become nA. The value of n is :

When a tuning fork (frequency 262 Hz) is struck, it loses half of its energy after 4s.The decay time ( τ ) and Q-factor for this tuning fork respectively are (Given, ln 2 = 0.693)

A number of holes are drilled along a diameter of a disc of radius R. To get minimum time period of oscillations, the disc should be suspended from a horizontal axis passing through a hole whose distance from the centre should be

A particle executes simple harmonic motion and is located at x = a , b and c at times t o , 3 t o and 5 t o respectively with respect to mean position . At t = 0 , the particle is at the mean position and moving toward positive x-axis. The frequency of the oscillation is

The amplitude of a damped oscillations decreases to 0.9 times its initial value in 5 seconds. By how many times to its initial value, energy of oscillation decreases in first 10 seconds?

Block A of mass 2kg moving with 10 m/s strikes a spring of constant π 2 N / m attached to another 2kg block B at rest kept on a smooth floor. The time for which rear moving block A remain in contact with spring will be

A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is 3.92 x 10 -3 m. What must be the least period of these oscillations, so that the object is not detached from the platform?

A object of certain mass is placed on a platform which oscillates up and down simple harmonically. If it is placed on the weighing machine, the reading of the machine will be

A liquid of density ρ contained in a vertical U tube filled to a height h on each limb. It executes simple harmonic motion in its limbs when a small disturbance is given. Then the time period of its oscillation is

The displacement of a particle is given by y = a sin 2 π t + ϕ then

The maximum acceleration of a particle in SHM is made twice keeping the maximum speed to be constant. It is possible when

Energy of a spring-mass oscillator increases if

If amplitude of simple harmonic motion doubles, its total energy becomes

s – t graph of a particle in SHM is as shown in figure. Choose the wrong option.

A solid sphere of mass m suspended by a inextensible massless string is given angular disturbance. The solid sphere is negatively charged. If it is allowed to oscillate above a large positively charged metallic plate. Its period

For the particle executing SHM of the displacement x is given by x = A cos ω t. Identify which represents variation of potential energy as a function of time t and displacement x.

A rod pivoted at one of its end undergoes simple harmonic motion when given a small oscillation. If the pivot is changed such that its moment of inertia decreases to half of its original value, then its new time period is

A particle is superposed with three SHMs simultaneously, in the same direction. Their time periods are 2 s, 3 s, 4s respectively. Then the time period of the resultant periodic motion is

The motion of a particle varies with time according to the relation y = 2 sin ω t + 2 cos ω t , then

A particle is superposed with two SHMs simultaneously, in the same direction. One has a time period 4 s and the other has a time period 5 s. What is the time period of the resultant periodic motion?

An object executes periodic motion which is given by sin 4 π t − c o s 4 π t . Then we can say that

Two. S.H.M’s are represented by the relations y = 10 sin 40 t + π 2 and y = 10 cos 25 t + π 4 . Then the ratio of their time periods is

Consider a body executes periodic motion given by 2 sin 2 t + π 4 m . Then we can concluded that

The displacement of a particle is represented by the equation y = 5 cos 3 π 2 − 2 π t where y is displacement ω is angular frequency in metre and t is given in instant of time in h the motion of the particle is

The time taken by a particle executing SHM of period T to move from the extreme position to half the maximum displacement is

The maximum acceleration of a body moving in SHM is a 0 and maximum velocity is v 0 . The amplitude is given by

A particle is undergoing SHM along a straight line so that its period is 12 s. The time it takes in traversing a distance equal to half its amplitude from the equilibrium position is

The velocity at the mean position of the bob of a pendulum is SHM is v. If its amplitude is doubled keeping the length the same, its velocity in the mean position will be

The maximum acceleration of a particle in SHM is made four times keeping the maximum speed to be constant. It is possible when

The maximum speed and acceleration of a particle executing simple harmonic motion are 10 c m s − 1 and 50 c m s − 2 . Magnitude of displacement of the particle when the speed is 2.5 c m s − 1 is

The equation of motion of a particle started at t = 0, is given by y = 5 sin 40 t + π 6 c m . The least time after which acceleration becomes maximum is

A slider is oscillating in SHM on an frictionless track with an amplitude A. You slow it so that its amplitude becomes half. Then the total mechanical energy in terms of previous value is

The displacement time graph of an oscillating particle. Find the maximum acceleration of the particle acceleration of the particle.

The x-t graph of a particle undergoing simple harmonic motion is shown below. The velocity of the particle at t = 4 3 sec is

The displacement time graph of an oscillating particle. Find the maximum velocity of the particle.

A solid cylinder of radius r and mass M rests on a curved path of radius R as shown figure. When displaced through a small angle θ as shown and left to itself, it executes simple harmonic motion. Then the time period of oscillation is (Assume that the cylinder rolls without slipping)

A particle executes SHM of time period T. The time taken by the particle to move from mean position to the position where its kinetic energy is equal to potential energy is

A particle of mass m is moving in a potential well, for which the potential energy is given by U ( x ) = U 0 1 − cos a x where U 0 and a are constants. Then for small oscillations, time period is

A simple pendulum with length L and mass m of the bob is oscillating with an amplitude A. The maximum tension in the string is

A body of mass 1kg executing S.H.M., its displacement y cm at t seconds is given by y = 6 sin π 100 π t + 1 4 . Its maximum kinetic energy is

A is a uniform solid cylinder, B is a uniform thin hollow cylinder. Both have same mass and same radius. C is a uniform plank. A, B and C each have same mass equal to m. Friction is sufficient so that there is no slipping anywhere. The time period of small oscillations of the above system is T= 2 π 15 m n K , when the plank is disturbed slightly. Here n is an integer. Find n.

A solid sphere of mass 6 kg and radius 20 cm is suspended from a wire. The time period of its angular oscillation if the torque constant of the wire is 2 × 10 − 3 N m / r a d is

An irregular lamina of mass 10 kg has moment of inertia 25 kg/m 2 is suspended from a steel wire of negligible mass whose couple per unit twist is 16 Nm/rad. It is allowed to execute small angular oscillations. It takes time T to complete one full oscillation. The value of T is

The x-t graph of a particle undergoing simple harmonic motion is shown below. The equation of the particle is given by

The displacement time graph of an oscillating particle is as shown in the diagram. The equation of motion of the particle is

The displacement of a particle varies with time as x = 12 sin ⁡ ω t − 16 sin 3 ⁡ ω t (in cm). If its motion is S.H.M., then its maximum acceleration is

A linear harmonic oscillator of force constant 2 × 10 6 N / m times and amplitude 0.01 m has a total mechanical energy of 160 joules. Its

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U ( x ) = k | x | 3 , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

A particle is performing simple harmonic motion along r-axis with amplitude 4 cm and time period 1 .2 sec. The minimum time taken by the particle to move from x=2 cm to x = + 4 cm and back again is given by

A particle executes simple harmonic motion (amplitude = A) between .x = -A and x = +A. The time taken for it to go from 0 to A/2 is T 1 and to go from A/2 to A is T 2 . Then

A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is 3.92 x 10 -3 m . What must be the least period of these oscillations, so that the object is not detached from the Platform?

The displacement y of a particle executing periodic motion is given by y = 4 cos 2 ⁡ ( t / 2 ) sin ⁡ ( 1000 t ) . This expression may be considered to be a result of the superposition of ……….. independent harmonic motions

The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination α , is given by

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 o , then

Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06m and force constant 0.1N/m. Initially both the balls are displaced by an angle θ = π / 6 radian with respect to the diameter PQ of the circle and released from rest. The frequency of oscillation of the ball B is

A particle of mass m is attached to three identical springs A, B and C each of force constant k as shown in figure. If the particle of mass m is pushed slightly against the spring A and released then the time period of oscillations is

The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is

Two pendulums have time periods T and 5T/4. They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has complete one oscillation?

As the expression is involving sine function, which of the following equations does not represent a simple harmonic motion?

A particle executing SHM of amplitude 4 cm and T = 4 sec. The time taken by it to move from positive extreme position to half the amplitude is

A man weighing 60 kg stands on the horizontal platform of a spring balance. The platform starts executing simple harmonic motion of amplitude 0.1 m and frequency 2 π Hz . Which of the following statement is correct?

Two particles P and Q describe SHM of same amplitude a, same frequency f along the same straight line. The maximum distance between the two particles is a 2 .The phase difference between the particle is

A particle executes SHM of amplitude A and time period T. The distance travelled by the particle in the duration its phase changes from π 12 to 5 π 12

The phase difference between the displacement and acceleration of a particle executing simple harmonic motion is

The equation of motion of a particle executing simple harmonic motion is a a + 16 π 2 x = 0 . In this equation, a is the linear acceleration in m/s 2 of the particle at a displacement x in meter. The time period in simple harmonic motion is

The phase difference between two particles executing SHM of the same amplitudes and frequency along same straight line while passing one another when going in opposite directions with equal displacement from their respective stating point is 2 π / 3 . If the phase of one particle is π / 6 , find the displacement at this instant, if amplitude is A

Time period (T) and amplitude (A) are same for two particles which undergo SHM along the same line. At one particular instant, one particle is at phase 3 π 2 and other is at phase zero. While moving in the same direction. Find the time at which they will cross each other:

Two simple harmonic motions are represented by equations y 1 = 4 sin ⁡ ( 10 t + ϕ ) y 2 = 5 cos ⁡ 10 t What is the phase difference between their velocities?

The diagram below shows a sinusoidal curve. The equation of the curve will be

Time period of a particle executing SHM is 8 s. At t = 0, it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is

A particle moves with a simple harmonic motion in a straight line. In the first second starting from rest it travels a distance a and in the next second it travels a distance b in the same direction. The amplitude of the motion is

The following figure shows the displacement versus time graph for two particle s A and B executing simple harmonic motions. The ratio of their maximum velocities is

A particle perfroms SHM with a period T and amplitude a. The mean velocity of the particle over the time interval during which it travels a distance a/2 from the extreme position is

A particle executing harmonic motion is having velocities v 1 and v 2 at distances is x 1 and x 2 from the equilibrium position. The amplitude of the motion is

A graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion is

The potential energy of a particle executing SHM along the x-axis is given by U = U 0 − U 0 cos ⁡ a x . What is the period of oscillation?

The variation of velocity of a particle executing SHM with time is shown in figure. The velocity of the particle when a phase change of π / 6 takes place from the instant it is at one of the extreme positions will be

A particle executing SHM of amplitude ‘a’ has a displacement a/2 at t = T/4 and a negative velocity. The epoch of the particle is

A body is executing Simple Harmonic Motion. At a displacement x its potential energy is E 1 and at a displacement y its potential energy is E 2 . The potential energy E at displacement (x +y) is

A vertical mass-spring system executes simple harmonic oscillations with a period of 2 s. A quantity of this system which exhibits simple harmonic variation with a period of 1 s is

A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement x. Which of the following statement is true?

A body at the end of a spring executes SHM with a period t 1 , while the corresponding period for another spring is t 2 . If the period of oscillation with the two spring in series is T, then

Two equal masses are suspended separately by two springs of constants k 1 and k 2 . If their oscillations satisfy the condition that their maximum velocities are equal, then the ratio of the amplitudes of the oscillations of the masses respectively is

Two identical springs are attached to a small block P. The other ends of the springs are fixed at A and B. When P is in equilibrium the extension of top spring is 20 cm and extension of bottom spring is 10 cm. The period of small vertical oscillations of P about its equilibrium position is (use g=9.8 m/s 2 )

A spring of spring constant k is cut into n equal parts, out of which r parts are placed in parallel and connected with mass M. The time period of oscillatory motion of mass M is

Two simple pendulums whose lengths are 100 cm and 121 cm are suspended side by side. Their bobs are pulled together and then released. After how many rninimum oscillations of the longer pendulum, will the two be in phase again

Three masses 700 g, 500 g and 400 g are suspended at the end of a spring a shown and are in equilibrium. When the 700 g mass is removed, the system oscillates with a period of 3 seconds, when the 500 g mass is also removed, it will oscillate with a period of

A simple pendulum hung from the ceiling of a train moving at constant speed has a period T. If the train starts accelerating or decelerating, then what will be the effect on time period of Pendulum?

Two simple pendulums of length 1 m and 16 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed n oscillations. The value of n is

Two pendulums of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is 5T/4, where T is the time period of shorter pendulum. Find the ratios of lengths of the two Pendulums.

A block of mass 1 kg hangs without vibrating at the end of a spring whose force constant is 200 N/m and which is attached to the ceiling of an elevator. The elevator is rising with an upward acceleration of g/3 when the acceleration suddenly ceases. The angular frequency of the block after the acceleration ceases is

A uniform square plate of side ‘a’ is hinged at one of its corners. It is suspended such that it can rotate about horizontal axis. The time period of small oscillation about its equilibrium position.

A particle is acted simultaneously by mutually perpendicular simple harmonic motion. x = a cos ⁡ ω t and y = a sin ⁡ ω t . The trajectory of motion of the particle will be

Four types of oscillatory systems: a simple pendulum; a physical pendulum; a torsional pendulum and a spring-mass system, each of same time period is taken to the Moon. If time periods are measured on the moon, which system or systems will have it unchanged?

A particle is subjected to two simple harmonic motions in the same direction having equal amplitude and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions, what is the phase difference between the two simple harmonic motions?

Three simple harmonic motions of equal amplitudes A and equal time periods along the same line combine. The phase of the second motion is 60 o ahead of the first and phase of the third motion is 60 o ahead of the second. The amplitude of resultant motion is

A wire is bent at an angle θ . A rod of mass m can slide along the bended wire without friction as shown in figure. A soap film is maintained in the frame kept in a vertical position and the rod is in equilibrium as shown in the figure. If rod is displaced slightly in vertical direction, then the time period of small oscillation of the rod is

One end of a spring of force constant k is fixed to a vertical wall and the other to a body of mass m resting on a smooth horizontal surface. There is another wall at a distance x 0 from the body. The spring is then compressed by 3x 0 and released. The time taken to strike the wall from the instant of release is (given sin − 1 ⁡ ( 1 / 3 ) = ( π / 9 )

A particle moves along a straight line to follow the equation a x 2 + b v 2 = k , where a, b and k are constants and x and v are x-coordinate and velocity of the particle respectively. Find the amplitude.

Two uniform ropes having linear mass densities m and 4m, length l. Each are joined to form a closed loop. The loop is hanging over a fixed frictionless small pulley with the lighter rope above as shown in the figure (in the figure equilibrium position is shown). Now if the point A joint, is slightly displaced in downward direction and released. It is found that, the loop perform angular SHM with the period of the oscillation equal to 2N. The value of N (take l = 150 π 2 metre) is .

Two springs with negligible masses and force constant of k 1 = 200 Nm − 1 and k 2 = 160 Nm − 1 are attached to the block of mass m = 10 kg as shown in the figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time t =0, sharp impulse of 50 Ns is given to the block with a hammer along the spring. Calculate δ if the amplitude of SHM is (5/ δ ) m.

Two simple pendulums A and B having lengths l and, l/4 respectively are released from the position as shown in figure. The time after which the release of the two strings become parallel for the first time is . [Take l = 90 π 2 a n d g = 10 m s – 2 ]

A spring is stretched by 0.2 m when a mass of 0.5 kg is suspended. The time period of the spring when a mass of 0.25kg is suspended and put to oscillation 2 π ψ s . The value of ψ (Given g = 9.8 ms -2 ) is .

A particle is moving on X-axis and has potential energy U = 2 − 20 x + 5 x 2 joule, where x is position. The particle is released at x = -3. If the mass of the particle is 0.1 kg, then the maximum velocity (in ms -1 ) of the particle is 25 β . If amplitude is 5 m, then the value of β is .

A sphere of mass M and radius R is on a smooth fixed inclined plane in equilibrium as shown in the figure. If now the sphere is displaced through a small distance along the plane, what will be the angular frequency (in rads -1 ) of the resulting SHM? Given, k = 4 M 3

A particle of mass 10 g is placed in- la- potential field given by V = 50 x 2 + 100 J/kg. If the frequency of oscillation is found to be x π cycle/s, the value of x is .

A uniform circular disc of radius R oscillates in a vertical plane about a horizontal axis, its minimum time period is π B 2 R g . The value of B is .

A rod of mass m and length L is pivoted at point O in a car whose acceleration towards left is a 0 . The rod is free to rotate in vertical plane. In equilibrium state, the rod remains horizontal when other end is suspended by a spring of force constant k. The time period of small oscillations of rod is T = 2 π D 3 . The value of D is . (Given, k = 20 Nm − 1 , a 0 = 10 ms − 2 , m = 1 kg , L = 1 m

The period of oscillations of mass m = 200 g poured into a bent tube (see figure), whose right arm makes an angle 30 o with the vertical and whose left arm is vertical, is T seconds. The cross-sectional area of the tube is A = 0.5 cm 2 . The value of 10T is . Neglect viscosity.

A particle is moving on x-axis has potential energy U = 2 − 20 x + 5 x 2 J along x-axis. The particle is released at x = – 3. The maximum value of ‘x’ is . [x is in meters and U is in joule]

A meter stick swinging in vertical plane about a fixed horizontal axis passing through its one end undergoes small oscillation of frequency f 0 . If the bottom half of the stick were cut off if its new frequency of small oscillation is found to be x f 0 . The value of x is .

A small ball is suspended by a thread of length l = 1 m at the point O on the wall, forming a small angle α = 2 ∘ with the vertical (as shown in figure). Then the thread with ball was deviated through a small angle β = 4 ∘ and set free. Assuming the collision of the ball against the wall to be perfectly elastic, the oscillation period of such a pendulum (in seconds) is (Take g = π 2 ).

The position of a particle with respect to origin varies according to the relation x = 3 sin ⁡ 100 t + 8 cos 2 ⁡ 50 t .The maximum displacement of the particle from the origin is .

A mass M attached to a spring oscillates with a period of 2 s. If the mass is increased by 2kg, the period increases by one second. The initial mass M (in kg) is .

Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The pendulum bob has mass 0.2 kg. The length of the pendulum in meter g = 10 m / s 2 is .

A particle of mass 0.8 kg is executing simple harmonic motion with an amplitude of 1.0 metre and periodic time 11/7 sec. The kinetic energy (in J) of the particle at the moment when it displacement is 0.6 metre is .

A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration 3 gm / s 2 . The period of small oscillations of the pendulum about its equilibrium position (in seconds) g = π 2 m / s 2 is .

A person normally weighing 60 kg stands on a platform which oscillates up and down harmonically at a frequency 2.0 s -1 and an amplitude 5.0 cm. If a machine on the platform gives the person’s weight against time, find the ratio of maximum and minimum reading it will show, take g = 10 m / s 2 .

In the arrangement shown in figure, pulleys ate small and light and springs are ideal K 1 = 25 π 2 , K 2 = 2 K 1 , K 3 = 3 K 1 and K 4 = 4 K 1 are force constants of the springs. The period of small vertical oscillations of block of mass k m=3kg is

A highly rigid cubical block B 1 of small mass m and side L is fixed rigidly onto another cubical block B 2 of same dimensions and of low modulus of rigidity η , such that the lower face of B 1 completely covers the upper face of B 2 . The lower face of B 2 is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of B 1 . After force is withdrawn, block B 1 executes small oscillations, the time period of which is π α m η L . The value of α is .

The time period of vertical oscillations of the system shown in Fig. is (given k 1 = k 2 = k and k 3 = 2 k )

A particle executing S.H.M. of amplitude 4 cm and T = 4 sec. The time taken by it to move from positive extreme position to half the amplitude is

Which one of the following is a simple harmonic motion

Two simple harmonic motions are represented by the equations y 1 = 0 .1 sin 100 πt + π 3 and y 2 = 0 .1 cosπt . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

Two particles are executing S.H.M. The equation of their motion are y 1 = 10 sin ω t + πT 4 , y 2 = 25 sin ω t + 3 πT 4 . What is the ratio of their amplitude

If x = asin ωt + π 6 and x ‘ = acosωt , then what is the phase difference between the two waves

A body is executing simple harmonic motion with an angular frequency 2 rad / s . The velocity of the body at 20 mm displacement, when the amplitude of motion is 60 mm, is

A particle executes S.H.M. with a period of 6 second and amplitude of 3 cm. Its maximum speed in cm/sec is

If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 sec, then its maximum velocity is

A particle has simple harmonic motion. The equation of its motion is x = 5 sin 4 t − π 6 , where x is its displacement. If the displacement of the particle is 3 units, then its velocity is

If the displacement of a particle executing SHM is given by y = 0 .30 sin ( 220 t + 0 .64 ) in metre, then the frequency and maximum velocity of the particle is

A particle is performing simple harmonic motion with amplitude A and angular velocity ω . The ratio of maximum velocity to maximum acceleration is

Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is

The velocity of a particle performing simple harmonic motion, when it passes through its mean position is

The velocity of a particle in simple harmonic motion at displacement y from mean position is

A particle executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.5 m/s 2 . The maximum velocity of the particle is (in m/s)

The maximum velocity of a simple harmonic motion represented by y = 3 sin 100 t + π 6 is given by

The instantaneous displacement of a simple pendulum oscillator is given by x = Acos ωt + π 4 . Its speed will be maximum at time

Which of the following is a necessary and sufficient condition for S.H.M.

The acceleration of a particle in S.H.M. is

The displacement of a particle moving in S.H.M. at any instant is given by y = asinωt . The acceleration after time t = T 4 is (where T is the time period)

The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

A small body of mass 0.10 kg is executing S.H.M. of amplitude 1.0 m and period 0.20 sec. The maximum force acting on it is

If a hole is bored along the diameter of the earth and a stone is dropped into hole

A body executing simple harmonic motion has a maximum acceleration equal to 24 metres / sec 2 and maximum velocity equal to 16 metres / sec . The amplitude of the simple harmonic motion is

For a particle executing simple harmonic motion, which of the following statements is not correct

A particle of mass 10 grams is executing simple harmonic motion with an amplitude of 0.5 m and periodic time of ( π / 5 ) seconds. The maximum value of the force acting on the particle is

The displacement of an oscillating particle varies with time (in seconds) according to the equation y (cm) = sin π 2 t 2 + 1 3 . The maximum acceleration of the particle is approximately

A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by Where A and K are positive constants

A particle executes harmonic motion with an angular velocity and maximum acceleration of 3.5 rad/sec and 7.5 m/s 2 respectively. The amplitude of oscillation is

In S.H.M. maximum acceleration is at

A body is vibrating in simple harmonic motion with an amplitude of 0.06 m and frequency of 15 Hz. The velocity and acceleration of body is

What is the maximum acceleration of the particle doing the SHM y = 2 sin πt 2 + φ where 2 is in cm

A 0.10 kg block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: x = ( 10 cm ) cos [ ( 10 rad / s ) t + π / 2 rad ] . What is the maximum acceleration experienced by the block

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion

Acceleration of a particle, executing SHM, at it’s mean position is

In simple harmonic motion, the ratio of acceleration of the particle to its displacement at any time is a measure of

A particle is executing simple harmonic motion with an amplitude of 0.02 metre and frequency 50 Hz. The maximum acceleration of the particle is

A particle executes simple harmonic motion along a straight line with an amplitude A. The potential energy is maximum when the displacement is

The total energy of a particle executing S.H.M. is proportional to

A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic

For a particle executing simple harmonic motion, the kinetic energy K is given by K = K o cos 2 ωt . The maximum value of potential energy is

The potential energy of a particle with displacement X is U(X). The motion is simple harmonic, when (K is a positive constant)

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = a) is

The total energy of the body executing S.H.M. is E. Then the kinetic energy when the displacement is half of the amplitude, is

The potential energy of a particle executing S.H.M. is 2.5 J, when its displacement is half of amplitude. The total energy of the particle be

The angular velocity and the amplitude of a simple pendulum is ω and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is

A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec and amplitude of 10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position is

The P.E. of a particle executing SHM at a distance x from its equilibrium position is

When the displacement is half the amplitude, the ratio of potential energy to the total energy is

For any S.H.M., amplitude is 6 cm. If instantaneous potential energy is half the total energy then distance of particle from its mean position is

A vertical mass-spring system executes simple harmonic oscillations with a period of 2 s. A quantity of this system which exhibits simple harmonic variation with a period of 1 s is

A body of mass 1 kg is executing simple harmonic motion. Its displacement y ( cm ) at t seconds is given by y = 6 sin ( 100 t + π / 4 ) . Its maximum kinetic energy is

A particle is executing simple harmonic motion with frequency f. The frequency at which its kinetic energy change into potential energy is

There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force F = − Kx , where x is the displacement. The total energy of body depends upon

The total energy of a particle executing S.H.M. is 80 J. What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position

In a simple harmonic oscillator, at the mean position

When a mass M is attached to the spring of force constant k, then the spring stretches by l. If the mass oscillates with amplitude l, what will be maximum potential energy stored in the spring

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)

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 displacement x. Which of the following statements is true

If <E> and <U> denote the average kinetic and the average potential energies respectively of mass describing a simple harmonic motion, over one period, then the correct relation is

The total energy of a particle, executing simple harmonic motion is

The kinetic energy of a particle executing S.H.M. is 16 J when it is at its mean position. If the mass of the particle is 0.32 kg, then what is the maximum velocity of the particle

A particle starts simple harmonic motion from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3 E / 4 . Its displacement at that instant is

A particle executes simple harmonic motion with a frequency f . The frequency with which its kinetic energy oscillates is

A particle of mass m is hanging vertically by an ideal spring of force constant K. If the mass is made to oscillate vertically, its total energy is

A body is moving in a room with a velocity of 20 m / s perpendicular to the two walls separated by 5 meters. There is no friction and the collisions with the walls are elastic. The motion of the body is

A tunnel has been dug through the centre of the earth and a ball is released in it. It will reach the other end of the tunnel after

The maximum speed of a particle executing S.H.M. is 1 m / s and its maximum acceleration is 1 .57 m / sec 2 . The time period of the particle will be

The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is

The acceleration of a particle performing S.H.M. is 12 cm / sec 2 at a distance of 3 cm from the mean position. Its time period is

If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to 2 .0 ms − 2 at any time, the angular frequency of the oscillator is equal to

Mark the wrong statement

A particle in SHM is described by the displacement equation x ( t ) = Acos ( ωt + θ ) . If the initial (t = 0) position of the particle is 1 cm and its initial velocity is π cm/s, what is its amplitude? The angular frequency of the particle is πs – 1

A particle executes SHM in a line 4 cm long. Its velocity when passing through the centre of line is 12 cm/s. The period will be

The displacement x (in metre) of a particle in, simple harmonic motion is related to time t (in seconds) as x = 0 .01 cos π t + π 4 The frequency of the motion will be

A simple harmonic wave having an amplitude a and time period T is represented by the equation y = 5 sinπ ( t + 4 ) m . Then the value of amplitude (a) in (m) and time period (T) in second are

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is

The displacement x (in metres) of a particle performing simple harmonic motion is related to time t (in seconds) as x = 0 .05 cos 4 π t + π 4 . The frequency of the motion will be

The period of a simple pendulum is doubled, when

The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is

A simple pendulum is made of a body which is a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off, the period of pendulum will

A pendulum suspended from the ceiling of a train has a period T, when the train is at rest. When the train is accelerating with a uniform acceleration a, the period of oscillation will

The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (If it is a second’s pendulum on earth)

A simple pendulum is set up in a trolley which moves to the right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle θ with the vertical

Which of the following statements is not true ? In the case of a simple pendulum for small amplitudes the period of oscillation is

The time period of a second’s pendulum is 2 sec. The spherical bob which is empty from inside has a mass of 50 gm. This is now replaced by another solid bob of same radius but having different mass of 100 gm. The new time period will be

A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration g / 4 , then the period of the pendulum will be

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by T = 2 π l g ‘ , where g ‘ is equal to

A second’s pendulum is placed in a space laboratory orbiting around the earth at a height 3R, where R is the radius of the earth. The time period of the pendulum is

The bob of a simple pendulum of mass m and total energy E will have maximum linear momentum equal to

The length of the second pendulum on the surface of earth is 1 m. The length of seconds pendulum on the surface of moon, where g is 1/6th value of g on the surface of earth, is

If the length of second’s pendulum is decreased by 2%, how many seconds it will lose per day

The period of simple pendulum is measured as T in a stationary lift. If the lift moves upwards with an acceleration of 5 g, the period will be

The length of a simple pendulum is increased by 1%. Its time period will

A simple pendulum with a bob of mass ‘m’ oscillates from A to C and back to A such that PB is H. If the acceleration due to gravity is ‘g’, then the velocity of the bob as it passes through B is

Identify correct statement among the following

The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle θ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

A simple pendulum executing S.H.M. is falling freely along with the support. Then

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of 60 o to the vertical, will be (If g = 10 m / s 2 )

If the metal bob of a simple pendulum is replaced by a wooden bob, then its time period will

A pendulum has time period T. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth then its time period on the other planet will be

A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21%, the percentage increase in the time period of the pendulum of increased length is

If the length of simple pendulum is increased by 300%, then the time period will be increased by

The length of a seconds pendulum is

A plate oscillated with time period ‘T’. Suddenly, another plate put on the first plate, then time period

A chimpanzee swinging on a swing in a sitting position, stands up suddenly, the time period will

A simple pendulum of length l has a brass bob attached at its lower end. Its period is T. If a steel bob of same size, having density x times that of brass, replaces the brass bob and its length is changed so that period becomes 2T, then new length is

If the length of the simple pendulum is increased by 44%, then what is the change in time period of pendulum

A simple pendulum, suspended from the ceiling of a stationary van, has time period T. If the van starts moving with a uniform velocity the period of the pendulum will be

To show that a simple pendulum executes simple harmonic motion, it is necessary to assume that

A simple pendulum consisting of a ball of mass m tied to a thread of length l is made to swing on a circular arc of angle θ in a vertical plane. At the end of this arc, another ball of mass m is placed at rest. The momentum transferred to this ball at rest by the swinging ball is

The ratio of frequencies of two pendulums are 2 : 3, then their lengths are in ratio

If the length of a pendulum is made 9 times and mass of the bob is made 4 times then the value of time period becomes

A pendulum of length 2m released at P. When it reaches Q, it losses 10% of its total energy due to air resistance. The velocity at Q is

Time period of a simple pendulum will be double, if we

The velocity of simple pendulum is maximum at

If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

Two bodies M and N of equal masses are suspended from two separate massless springs of force constants k 1 and k 2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude M to that of N is

A spring has a certain mass suspended from it and its period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is now

In arrangement given in figure, if the block of mass m is displaced, the frequency is given by

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are K 1 and K 2 . The time period of the suspended mass will be

A spring is stretched by 0.20 m, when a mass of 0.50 kg is suspended. When a mass of 0.25 kg is suspended, then its period of oscillation will be ( g = 10 m / s 2 )

A weightless spring which has a force constant oscillates with frequency n when a mass m is suspended from it. The spring is cut into two equal halves and a mass 2m is suspended from it. The frequency of oscillation will now become

A particle of mass 200 gm executes S.H.M. The restoring force is provided by a spring of force constant 80 N / m. The time period of oscillations is

A uniform spring of force constant k is cut into two pieces, the lengths of which are in the ratio 1 : 2. The ratio of the force constants of the shorter and the longer pieces is

A block of mass m, attached to a spring of spring constant k, oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. The block has a speed v when the spring is at its natural length. Before coming to an instantaneous rest, if the block moves a distance x from the mean position, then

The force constants of two springs are K 1 and K 2 . Both are stretched till their elastic energies are equal. If the stretching forces are F 1 and F 2 , then F 1 : F 2 is

If the period of oscillation of mass m suspended from a spring is 2 sec, then the period of mass 4m will be

Five identical springs are used in the following three configurations. The time periods of vertical oscillations in configurations (i), (ii) and (iii) are in the ratio

What will be the force constant of the spring system shown in the figure

Two springs have spring constants K A and K B and K A > K B . The work required to stretch them by same extension will be

The effective spring constant of two spring system as shown in figure will be

A mass M is suspended by two springs of force constants K 1 and K 2 respectively as shown in the diagram. The total elongation (stretch) of the two springs is

The scale of a spring balance reading from 0 to 10 kg is 0.25 m long. A body suspended from the balance oscillates vertically with a period of π / 10 second. The mass suspended is (neglect the mass of the spring)

If a spring has time period T, and is cut into n equal parts, then the time period of each part will be

One-fourth length of a spring of force constant K is cut away. The force constant of the remaining spring will be

A mass m is suspended separately by two different springs of spring constant K 1 and K 2 gives the time-period t 1 and t 2 respectively. If same mass m is connected by both springs as shown in figure then time-period t is given by the relation

The springs shown are identical. When A=4kg , the elongation of spring is 1 cm. If B=6kg, the elongation produced by it is

If a spring extends by x on loading, then energy stored by the spring is (if T is the tension in the spring and K is the spring constant)

A weightless spring of length 60 cm and force constant 200 N/m is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of 0.25 kg is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes S.H.M. of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of m/M is

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15 cm/sec and the period is 628 milli-seconds. The amplitude of the motion in centimeters is

When a mass m is attached to a spring, it normally extends by 0.2 m. The mass m is given a slight addition extension and released, then its time period will be

A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is f 1 . The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is f 2 . Which of the following relations between the frequencies is correct

A mass m oscillates with simple harmonic motion with frequency f = ω 2 π and amplitude A on a spring with constant K , therefore

Two masses m 1 and m 2 are suspended together by a massless spring of constant K. When the masses are in equilibrium, m 1 is removed without disturbing the system. The amplitude of oscillations is

The motion of a particle varies with time according to the relation y = a ( sinω t + cosω t ) , then

The displacement of a particle varies according to the relation x = 4(cos π t + sin π t). The amplitude of the particle is

In case of a forced vibration, the resonance wave becomes very sharp when the

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U ( x ) = k [ x ] 3 , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of gas. The cylinder is kept with its axis horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be

A sphere of radius r is kept on a concave mirror of radius of curvature R. The arrangement is kept on a horizontal table (the surface of concave mirror is frictionless and sliding not rolling). If the sphere is displaced from its equilibrium position and left, then it executes S.H.M. The period of oscillation will be

The amplitude of vibration of a particle is given by a m = ( a 0 ) / ( aω 2 − bω + c ) ; where a 0 , a , b and c are positive. The condition for a single resonant frequency is

A U tube of uniform bore of cross-sectional area A has been set up vertically with open ends facing up. Now m gm of a liquid of density d is poured into it. The column of liquid in this tube will oscillate with a period T such that

A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 sec. The minimum time taken by the particle to move from x =2 cm to x = + 4 cm and back again is given by

A large horizontal surface moves up and down in SHM with an amplitude of 1 cm. If a mass of 10 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of S.H.M. will be

Due to some force F 1 a body oscillates with period 4/5 sec and due to other force F 2 oscillates with period 3/5 sec. If both forces act simultaneously, the new period will be

A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is 3 .92 × 10 − 3 m . What must be the least period of these oscillations, so that the object is not detached from the platform

A particle executes simple harmonic motion (amplitude = A) between x = − A and x = + A . The time taken for it to go from 0 to A/2 is T 1 and to go from A/2 to A is T 2 . Then

Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed …. oscillations.

The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0 sec, the tension in the string when the bob passes through O is

The metallic bob of a simple pendulum has the relative density ρ . The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by

A clock which keeps correct time at 20 o C , is subjected to 40 o C . If coefficient of linear expansion of the pendulum is 12 × 10 − 6 / ° C . How much will it gain or loose in time

The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination θ , is given by

A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω 0 -An external force F (t) proportional to cosω t ( ( ω ≠ ω 0 ) is applied to the oscillator. The time displacement of the oscillator will be proportional to

An ideal spring with spring-constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is

The displacement y of a particle executing periodic motion is given by y = 4 cos 2 ( t / 2 ) sin ( 1000 t ) . This expression may be considered to be a result of the superposition of ……….. independent harmonic motions

The function sin 2 ( ωt ) represents

A simple pendulum is hanging from a peg inserted in a vertical wall. Its bob is stretched in horizontal position from the wall and is left free to move. The bob hits on the wall the coefficient of restitution is 2 5 . After how many collisions the amplitude of vibration will become less than 60°

A brass cube of side a and density σ is floating in mercury of density ρ . If the cube is displaced a bit vertically, it executes S.H.M. Its time period will be

Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06 π m and force constant 0.1N/m. Initially both the balls are displaced by an angle θ = π / 6 radian with respect to the diameter PQ of the circle and released from rest. The frequency of oscillation of the ball B is

A hollow sphere is filled with water through a small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flows out of the hole at the bottom, the period of oscillation will

A particle of mass m is attached to three identical springs A, B and C each of force constant k a shown in figure. If the particle of mass m is pushed slightly against the spring A and released then the time period of oscillations is

A uniform rod of length 2.0 m is suspended through an end and is set into oscillation with small amplitude under gravity. The time period of oscillation is approximately

The acceleration a of a particle undergoing S.H.M. is shown in the figure. Which of the labelled points corresponds to the particle being at – x max

The graph shows the variation of displacement of a particle executing S.H.M. with time. We infer from this graph that

For a particle executing S.H.M. the displacement x is given by x = Acosωt . Identify the graph which represents the variation of potential energy (P.E.) as a function of time t and displacement x

A body of mass 0.01 kg executes simple harmonic motion (S.H.M.) about x = 0 under the influence of a force shown below : The period of the S.H.M. is

The variation of the acceleration a of the particle executing S.H.M. with displacement y is as shown in the figure

The variation of potential energy of harmonic oscillator is as shown in figure. The spring constant is

A body performs S.H.M. Its kinetic energy K varies with time t as indicated by graph

In a certain observation of experiment with simple pendulum we get l = 23 . 2 cm , r = 1 . 32 cm and time taken for 20 oscillations was 20 . 0 s e c . Taking π 2 = 10 , find the value of g in proper significant figures.

For different values of L, we get different values of T 2 . The graph between L versus T 2 is as shown in figure. Find the value of ‘ g ‘ from the given graph. Take ( π 2 = 10 .

Why should the amplitude be small for a simple pendulum experiment?

What type of graph do you expect between L a n d T .

Does the time period depend upon the mass and the material of the bob?

What type of graph do you expect between L and T 2 ?

Why do the pendulum clocks go slow in summer and fast in winter?

The second’s pendulum is taken from earth to moon, to keep the time period constant

A simple pendulum has a bob which is a hollow sphere full of sand and oscillates with certain period. If all that sand is drained out through a hole at its bottom, then its period

Why do we use Invar material for the pendulum of good clocks?

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 min -1 , its maximum speed is

Which of the following motions is not simple harmonic?

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency (in Hz) of vibration of the 8 kg block is .

Two blocks each of mass m is connected to the spring of spring constant 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

Physics – Oscillations Questions for CBSE Class 11th

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