CBSE Class 12 Physics Simple Harmonic Motion Assignment

Question. A mass m = 100 gms is attached at the end of a light spring which oscillates on a friction less horizontal table with an amplitude equal to 0.16 meter and the time period equal to 2 sec. Initially the mass is released from rest at t = 0 and displacement x = – 0.16 meter. The expression for the displacement of the mass at any time (t) is:
a. x = 0.16cos (π t)
b. x = −0.16cos(π t)
c. x = 0.16cos(π t +π )
d. x = −0.16cos(π t +π )
Answer : B

Question. A particle of mass 10 grams is executing S.H.M. with an amplitude of 0.5 meter and circular frequency of 10 radian/sec. The maximum value of the force acting on the particle during the course of oscillation i:s
a. 25 N
b. 5 N
c. 2.5 N
d. 0.5 N
Answer : D

Question. The equation of motion of a particle is d²y/dt² + ky = 0 where k is a positive constant. The time period of the motion is given by:
a. 2π/k
b. 2πk
c. 2π/k
d. 2π√k
Answer : C

Question. The motion of a particle executing S.H.M. is given by x = 0.01 sin100π (t + .05) . Where x is in meter and time t is in seconds. The time period is:
a. 0.01 sec
b. 0.02 sec
c. 0.1 sec
d. 0.2 sec
Answer : B

Question. A particle perform simple harmonic motion. The equation of its motion is x = 5sin (4t - π/6)  Where x is its displacement. If the displacement of the particle is 3 units then its velocity is:
a. 2π/3
b. 5π/6
c. 20
d. 16
Answer : D

Question. A body is executing simple harmonic motion with an angular frequency 2 rad/sec. The velocity of the body at 20 mm displacement. When the amplitude of motion is 60 mm is:
a. 40 mm/sec
b. 60 mm/sec
c. 113 mm/sec
d. 120 mm/sec
Answer : C

Question. 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. 35 Hz, 66 m / s
b. 45 Hz, 66 m / s
c. 58 Hz, 113 m / s
d. 35 Hz, 132 m / s
Answer : A

Question. The amplitude and the time period in a S.H.M. is 0.5 cm and 0.4 sec respectively. If the initial phase is π / 2 radian, then the equation of S.H.M. will be:
a. y = 0.5 sin 5πt
b. y = 0.5 sin 4πt
c. y = 0.5 sin 2.5πt
d. y = 0.5 cos 5πt
Answer : D

Question. A particle in S.H.M. is described by the displacement function x(t) = a cos(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. 1 cm
b. 2 cm
c. 2 cm
d. 2.5 cm
Answer : B

Question. 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:
a. 1 sec
b. 1/3 sec
c. 2/3 sec
d. √3 / 2 sec'
Answer : C

Question. 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√3cm / s, will be:
a. 2√3cm
b. √3cm
c. 1 cm
d. 2 cm
Answer : D

Question. vTwo simple harmonic motions are represented by the equations y₁ = 0.1 sin (100πt + π/3) y₂ = 0.1 cos πt.
The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is:
a. -π/3
b. π/6
c. -π/6
d. π/3
Answer : C

Question. 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. πA√3/3
b. πA/T
c. πA√3/2T
d. 3π2A/T
Answer : A

Question. 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. 5
b. 5√2
c. 5√3
d. 10√2
Answer : C

Question. 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. πA √3/T
b. πA/T
c. πA√3/2T
d. 3π²A/TA
Answer : A

Question. If x = a sin (ωt + π/6) and x ′ = a cosωt , then what is the phase difference between the two waves:
a. π/3
b. π/6
c. π/2
d. π
Answer : A

Question. 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 collision with the walls are elastic. The motion of the body is:
a. Not periodic
b. Periodic but not simple harmonic
c. Periodic and simple harmonic
d. Periodic with variable time period
Answer : B

Question. The instantaneous displacement of a simple pendulum oscillator is given by a = A cos
(ωt +  π/4). Its speed will be maximum at time:
a. π/4ω
b.π'/2ω
c . π/ω
d 2/ ω
Answer :  A

Question. 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:
a. 5.65 m/s and 5.32 × 10² m/s²
b. 6.82 m/s and 7.62 × 10² m/s²
c. 8.91m/s and 8.21 × 10² m/s²
d. 9.82 m/s and 9.03 × 10² m/s²
Answer : A

Question. The velocity of a particle performing simple harmonic motion, when it passes through its mean position is:
a. Infinity
b. Zero
c. Minimum
d. Maximum
Answer : D

Question. The displacement of a particle moving in S.H.M. at any instant is given by y = a sinωt . The acceleration after time t = 4/T is: (where T is the time period)
a. aω
b. – aω
c. aω²
d. – aω²
Answer : D

Question. For a particle executing simple harmonic motion, which of the following statements is not correct?
a. The total energy of the particle always remains the same
b. The restoring force of always directed towards a fixed point
c. The restoring force is maximum at the extreme positions
d. The acceleration of the particle is maximum at the equilibrium position
Answer : D

Question. 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. 100 m/s²
b. 100 π² m/s²
c. 100 m/s²
d. 200 π² m/s²
Answer : D

Question. 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:
a. 1/2π√3
b. 2π√3
c. 2π/√3
d. √3/2π
Answer : A

Question. The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = a) is:
a. a/2
b. a√2
c. a/√2
d. a√2/3
Answer : B

Question. 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:
a. 37.5π² erg
b. 3.75π² erg
c. 375π² erg
d. 0.375π² erg
Answer : C

Question. 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:
a. 84.6 minutes
b. 42.3 minutes
c. 1 day
d. Will not reach the other end
Answer : A

Question. 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?
a. P.E. is maximum when x = 0
b. K.E. is maximum when x = 0
c. T.E. is zero when x = 0
d. K.E. is maximum when x is maximum
Answer : B

Question. A particle executes simple harmonic motion with a frequency f . The frequency with which its kinetic energy oscillates is:
a. f / 2
b. f
c. 2 f
d. 4 f
Answer : C

Question. A particle in SHM is described by the displacement equation x(t) = A cos(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. 1 cm
b. √2 cm
c. 2 cm
d. 2.5 cm
Answer : D

Question. When the displacement is half the amplitude, the ratio of potential energy to the total energy is:
a. 1/2
b. 1/4
c. 1
d. 1/8
Answer : B

Question. 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:
a. K, x
b. K, a
c. K, a, x
d. K, a, v
Answer : B

Question. 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. a = 10, T = 2
b. a = 5, T = 1
c. a = 10 , T = 1
d. a = 5, T = 2
Answer : C

Question. 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:
a. (π/5)sec
b. 2π sec
c. 20π sec
d. 5π sec
Answer : D

Question. To make the frequency double of an oscillator, we have to:
a. Double the mass
b. Half the mass
c. Quadruple the mass
d. Reduce the mass to one-fourth
Answer : B

Question. 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:
a. E/2
b. E/4
c. 3E/4
d. √3E/4
Answer : C

Question. The kinetic energy and potential energy of a particle executing S.H.M. will be equal, when displacement is: (amplitude = a)
a. a/2
b. a√2
c. a/√2
d. a√2/3
Answer : B

Question. 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. 4 sec
b. 1 sec
c. 2 sec
d. 8 sec
Answer : A

Question. The length of a simple pendulum is increased by 1%. Its time period will:
a. Increase by 1%
b. Increase by 0.5%
c. Decrease by 0.5%
d. Increase by 2%
Answer : D

Integer

Question. In damped oscillations, the amplitude of oscillations is reduced to one-third of its initial value of 9 cm at end of 100 oscillations. What will be its amplitude of oscillation in cm when it completes 200 oscillations?
Answer : 1

Question. A simple pendulum with a brass bob has a time period 2√2 second. The bob is now immersed in a non viscous liquid and oscillated. If the density of liquid is (1/9) that of brass, find the time period in second of the same pendulum?
Answer : 3

Question. A mass m attached to a spring oscillates with a period of 2 seconds. If the mass is increased by 3 kg. the period increases by 2 seconds. Find the initial mass M in kg, assuming that Hooke’s law is obeyed.
Answer : 1

Question. A particle is performing SHM along x-axis with amplitude M4.0 cm and time period 1.2 s. What is the minimum time in deci-second taken by the particle to move from x = +2 cm to x = + 4 cm and back again?
Answer : 4

Question. A light pointer fixed to one prong of a tuning fork touches gently a smoked vertical plate. The fork is set vibrating and the plate is allowed to fall freely. Two complete oscillations are traced when the plate falls through 40 cm.
What is the frequency (inHz) of the tuning fork?
Answer : 7

Question. A body of mass 1kg is executing simple harmonic motion. Its displacement y(cm) at t seconds is given by y = 6 (100t + π/4) . Its maximum kinetic energy is
Answer : 18

Question. The maximum speed of a particle executing S.H.M. is 1m / s and its maximum acceleration is 1.87 m/sec². The time period of the particle will be
Answer : 4

Question. The S.H.M. of a particle is given by the equation y = 3 sinω t + 4 cosω t . The amplitude is
Answer : 5

Question. A particle moves according to law x = a cosπ t. The distance covered by it in 2.5 second is:
Answer : 5

Question. The time period of a seconds’ pendulum is 2 sec. The spherical bob which is empty from inside has a mass 50 g. This is now replaced by another solid bob of same radius but having a different mass of 1000 g. The new time period will be:
Answer : 2

Comprehension-1

When force acting on the particle is of nature F = – kx, motion of particle is S.H.M. Velocity at extreme is zero while at mean position it is maximum. In case of acceleration situation is just reverse. Maximum displacement of particle from mean position on both sides is same and is known as amplitude.Refer to figure. One kg block performs vertical harmonic oscillations with amplitude 1.6 cm and frequency 25 rad s–1.

1. The maximum value of the force that the system exerts on the surface is
(a) 20 N
(b) 30 N
(c) 40 N
(d) 60 N

2. The minimum force is
(a) 20 N (b) 30 N (c) 40 N (d) 60 N

Each of the questions given below consists of two statements, an assertion (A) and reason (R).
Select the number corresponding to the appropriate alternative as follows
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

3. A: All periodic motions are oscillatory.
R: Oscillating system do not remains confined within two extreme limits.

4. A: The time period of a simple pendulum inside a lift falling freely is zero.
R: A body falling freely has infinite acceleration.

5. A. A body is oscillating, about its mean position O, between two extreme positions P and Q and with amplitude A. After crossing O in the positive (right) direction, when it is at C, the mid-point between O and Q, its phase is π/6 radian or 30º.]

R. Phase of a particle, executing SHM is directly proportional to its displacement from the mean
position.

6. A. When a spring is cut into three exactly equal parts, then force constant of each smaller part is three times the force constant of the original spring.
R. Force constant is inversely proportional to the length

7. A. A state of stable equilibrium is required for SHM.
R. Restoring force always acts in the opposite direction to displacement.

Assignment
1. (d) 2. (c) 3. (d) 4. (d) 5. (c)   6. (a) 7. (a)