WBCHSE Class 11 Physics MCQS – Simple Harmonic Motion

Simple Harmonic Motion Multiple Choice Question And Answers

Question 1. The motion of any hand of a clock is a (an)

1. Periodic motion
2. Simple harmonic motion
3. Vibration
4. Oscillation

Answer: 1. Periodic motion

Question 2. if the mass of a particle executing SHM is m and its angular frequency is ω, then the force constant of that SHM will be

1. \(m \omega\)
2. \(m \omega^2\)
3. \(\sqrt{\frac{m}{\omega}}\)
4. \(\sqrt{\frac{\omega}{m}}\)

Answer: 2. \(m \omega^2\)

Question 3. In the case of a simple harmonic motion, which of the following statements is not true?

1. The moving particle repeats the same path periodically
2. The restoring force acting on the particle is always directed toward the equilibrium position
3. The restoring force acting on the particle is always proportional to its displacement
4. The restoring force acting on the particle is always proportional to the velocity of the particle

Answer: 4. The restoring force acting on the particle is always proportional to the velocity of the particle

Question 4. If the displacement and the restoring force acting on a particle executing simple harmonic motion are x and F respectively, then F = – kx. Here the negative sign on the right-hand side indicates that

1. The restoring force is directed toward the equilibrium position
2. The restoring force is directly proportional to the displacement
3. The force constant is always negative
4. The restoring force is always negative

Answer: 1. The restoring force is directed toward the equilibrium position

Question 5. If the mass of a particle executing SHM is m and its angular frequency is ω, then the period of its oscillation will be

1. \(\frac{1}{\omega}\)
2. \(\frac{m}{\omega}\)
3. \(\frac{\omega}{2 \pi}\)
4. \(\frac{2 \pi}{\omega}\)

Answer: 4. \(\frac{2 \pi}{\omega}\)

Question 6. The magnitude of the maximum velocity of the SHM expressed by the equation x – Asinωt is

1. A
2. Aω
3. Aω²
4. A²ω

Answer: 2.

Question 7. The magnitude of maximum acceleration of the SHM expressed by the equation x = A sinωt is

1. A
2. Aω
3. Aω²
4. A²ω

Answer: 3. Aω²

Question 8. If the equation x = asinωt represents a simple harmonic motion of a particle, then its initial position is

1. Equilibrium point
2. Terminal point
3. Any point in the right side of the point of equilibrium
4. Any point in the left side of the point of equilibrium

Answer: 1. Equilibrium point

Question 9. The time period of the SHM expressed by the equation x = 4sin4πt is

1. 4s
2. 4πs
3. 2s
4. \(\frac{1}{2}\)s

Answer: 4. \(\frac{1}{2}\)s

Question 10. If the displacement and the acceleration of a particle executing SHM at any instant are x and a respectively, then the time period of that motion will be

1. \(2 \pi \sqrt{\frac{a}{x}}\)
2. \(2 \pi \sqrt{\frac{x}{a}}\)
3. \(\frac{1}{2 \pi} \sqrt{\frac{a}{x}}\)
4. \(\frac{1}{2 \pi} \sqrt{\frac{x}{a}}\)

Answer: 2. \(2 \pi \sqrt{\frac{x}{a}}\)

Question 11. Which one of the following is not the equation of an SHM?

1. F = -kx
2. a = –\(\frac{k}{m}\)x
3. a = -ω²x
4. a = ω²x

Answer: 4. a = ω²x

Question 12. A particle executing SHM follows a straight path of length l. The amplitude of its motion is

1. 2l
2. l
3. \(\frac{l}{2}\)
4. \(\frac{l}{4}\)

Answer: \(\frac{l}{2}\)

Question 13. The phase difference between two SHMs x = Bcosωt and x = Asinωt is

1. 180°
2. 90°
3. -90°
4. Zero

Answer: 2. 90°

Question 14. The amplitude of vibration of the SHM represented by the equation x = A sinωt+ B cosωt is

1. A+B
2. A-B
3. \(\sqrt{A^2+B^2}\)
4. \(\sqrt{A^2-B^2}\)

Answer: 3. \(\sqrt{A^2+B^2}\)

Question 15. The initial phase or epoch of the SHM represented by the equation x = A sinωt+ B cosωt is

1. \(\frac{A}{B}\)
2. \(\frac{B}{A}\)
3. \(\tan ^{-1} \frac{A}{B}\)
4. \(\tan ^{-1} \frac{B}{A}\)

Answer: 4. \(\tan ^{-1} \frac{B}{A}\)

Question 16. The SHM executed by a particle of mass 2 kg is represented by the equation x = 4sin4πtm. Total mechanical energy of the particle (in joule) will be

1. 256π²
2. 6π²
3. 16π²
4. 16π

Answer: 1. 256π²

Question 17. A particle is executing SHM with frequency a. The frequency of the variation of its kinetic energy is

1. \(\frac{a}{2}\)
2. a
3. 2a
4. 4a

Answer: 3. 2a

Question 18. The distance between the positions of maximum potential energy and maximum kinetic energy of a particle executing SHM is

1. \(\pm \frac{A}{2}\)
2. \(\pm \frac{A}{\sqrt{2}}\)
3. \(\pm A\)
4. \(\pm 2 A\)

Answer: 3. \(\pm A\)

Question 19. If the amplitude of an SHM is A, then for what position of the particle, half of its total energy will be potential energy and the remaining half will be kinetic energy?

1. \(\pm \frac{A}{2}\)
2. \(\pm \frac{A}{\sqrt{2}}\)
3. \(\pm \frac{A}{3}\)
4. \(\pm \frac{A}{2 \sqrt{2}}\)

Answer: 2. \(\pm \frac{A}{\sqrt{2}}\)

Question 20. Kinetic energy and potential energy of a simple harmonic motion are K and V respectively. Then which one is always true

1. K>V
2. K<V
3. K = V
4. K+V = constant

Answer: 4. K+V = constant

Question 21. When a spring is stretched by 3 cm, stored potential energy becomes u and when it is stretched by 6 cm potential energy becomes

1. 2u
2. 3u
3. 4u
4. 6u

Answer: 3. 4u

Question 22. The time period of a simple pendulum is 2 s. If its length is doubled, then the new time period will be

1. 2s
2. √2 s
3. 2√2 s
4. 4s

Answer: 3. 2√2 s

Question 23. If the time period of a simple pendulum of effective length L is T, then the effective length of a simple pendulum having time period 2 T will be

1. \(\frac{L}{2}\)
2. L
3. 2 L
4. 4L

Answer: 4. 4L

Question 24. The time period of a second pendulum is

1. 1s
2. 2s
3. \(\frac{1}{2}\)s
4. \(\frac{1}{2}\)s

Answer: 2. 2s

Question 25. If a second pendulum is taken to the surface of the moon from the earth, its time period would be (acceleration due to gravity on the surface of the moon is 1/6th that on the earth’s surface)

1. 12 s
2. 6 s
3. 26 s
4. \(\frac{2}{\sqrt{6}}\)s

Answer: 3. 26 s

Question 26. The nature of the graph of the effective length of a pendulum versus its time period will be

1. Linear
2. Parabolic
3. Exponential
4. Sinusoidal

Answer: 2. Parabolic

Question 27. The length of a second pendulum on the surface of the earth is

1. 1 m (approx.)
2. 1.1m (approx.)
3. 0.25m (approx.)
4. 2m (approx.)

Answer: 1. 1 m (approx.)

Question 28. The time period of a simple pendulum on the surface of the earth is T₁ and at a height R above the surface of the earth is T₂; where R is the radius of the earth. The ratio T₁/T₂ is

1. 1
2. √2
3. 4
4. 2

Answer: 4. 2

Question 29. A simple pendulum has time period T₁. The point of suspension is now moved upwards according to the relation y = Kt² (K=1m · s^-2) where y is the vertical displacement. The time period now becomes T₂ . The ratio of \(T_1^2 / T_2^2 \text { is }\left(\mathrm{g}=10 \mathrm{~m} \cdot \mathrm{s}^{-2}\right)\)

1. \(\frac{6}{5}\)
2. \(\frac{5}{6}\)
3. 1
4. \(\frac{4}{5}\)

Answer: 1. \(\frac{6}{5}\)

Question 30. The length of a pendulum is l. The bob is pulled to one side to make an angle with the vertical and is then released. The velocity of the bob, when it crosses the position of equilibrium, is

1. \(\sqrt{2 g l}\)
2. \(\sqrt{2 g l \cos \alpha}\)
3. \(\sqrt{2 g l(1-\cos \alpha)}\)
4. \(\sqrt{2 g l(1-\sin \alpha)}\)

Answer: 3. \(\sqrt{2 g l(1-\cos \alpha)}\)

Question 31. A simple pendulum of length l has a maximum angular displacement θ. The maximum kinetic energy of the bob of mass m will be

1. mgl(1-cosθ)
2. mgl cosθ
3. mgl sinθ
4. None of these

Answer: 1. mgl(1-cosθ)

Question 32. The mass M shown oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is

1. \(\frac{k_1 A}{k_2}\)
2. \(\frac{k_2 A}{k_1}\)
3. \(\frac{k_1 A}{k_1+k_2}\)
4. \(\frac{k_2 A}{k_1+k_2}\)

Answer: 4. \(\frac{k_2 A}{k_1+k_2}\)

Question 33. A body of mass M is held between two massless springs, on a smooth inclined plane as shown. The free ends of the springs are attached to firm supports. If each spring has force constant k, the period of oscillation of the body is

1. \(2 \pi \sqrt{\frac{M}{2 k}}\)
2. \(2 \pi \sqrt{\frac{2 M}{k}}\)
3. \(2 \pi \sqrt{\frac{M g \sin \theta}{2 k}}\)
4. \(2 \pi \sqrt{\frac{M g \sin \theta}{k}}\)

Answer: 1. \(2 \pi \sqrt{\frac{M}{2 k}}\)

Question 34. A wooden cube (density of wood d) of side l floats in a liquid of density ρ with its upper and lower surfaces horizontal. lf the cube is pushed slightly down and released, and it performs simple harmonic motion of period T, then T is equal

1. \(2 \pi \sqrt{\frac{l \rho}{(\rho-d) g}}\)
2. \(2 \pi \sqrt{\frac{l \rho}{d g}}\)
3. \(2 \pi \sqrt{\frac{l d}{\rho g}}\)
4. \(2 \pi \sqrt{\frac{l d}{(\rho-d) g}}\)

Answer: \(2 \pi \sqrt{\frac{l \rho}{d g}}\)

Question 35. A spring is cut into two pieces in such a way that one piece is double the length of the other. If the force constant of the main spring is k then fore constant of the longer part is

1. \(\frac{2}{3}k\)
2. \(\frac{3}{2}k\)
3. 3k
4. 6k

Answer: 2. \(\frac{3}{2}k\)

In this type of question, more than one option is correct.

Question 36. The displacement-time relation for a particle can be expressed as x = 0.5 [cos²(nπt)-sin²(nπt)]. This relation shows that

1. The particle executes SHM with an amplitude 0.5 m
2. The particle executes SHM with a frequency n times that of a second pendulum
3. The particle executes SHM and the velocity in its mean position is (3.142 n)m • s^-1
4. The article does not execute SHM at all

Answer:

1. The particle executes SHM with amplitude of 0.5 m

3. The particle executes SHM and the velocity in its mean position is (3.142 n)m • s^-1

Question 37. A simple pendulum consists of a bob of mass m and a light string of effective length L as shown,

Another identical ball moving with a small velocity v₀ collides with the pendulum’s bob and sticks to it. For this new pendulum of mass 2m, mark the correct statement(s).

1. Time period of the pendulum is \(2 \pi \sqrt{\frac{L}{g}}\)
2. The equation of motion for this pendulum is \(\theta=\frac{v_0}{2 \sqrt{g L}} \sin \left[\sqrt{\frac{g}{L}} t\right]\)
3. The equation of motion for this pendulum is \(\theta=\frac{v_0}{2 \sqrt{g L}} \cos \left[\sqrt{\frac{g}{L}} t\right]\)
4. Time period of the pendulum is \(2 \pi \sqrt{\frac{2 L}{g}}\)

Answer:

Question 38. The function, x = A sin²ωt + B cos²ωt+ C sinωtcosωt represents SHM

1. For any value of A, B, and C (except C = 0)
2. If A = -B, C = 2B, amplitude = |B√2|
3. If A = B; C = 0
4. If A = B; C = 2B, amplitude = |B|

Answer:

2. If A = -B, C = 2B, amplitude = |B√2|

4. If A = B; C = 2B, amplitude = |B|

Question 39. Choose the correct statement(s).

1. The time period of the spring-mass system will change when it is made to oscillate horizontally and vertically.
2. Natural frequency depends upon elastic properties and dimensions of the body.
3. At the mean position, the energy is entirely KE, and at extreme positions, the energy is entirely potential.
4. A pendulum having a time period of 2 seconds is called a second’s pendulum.

Answer:

2. Natural frequency depends upon elastic properties and dimensions of the body.

3. At the mean position, the energy is entirely KE and at extreme positions, the energy is entirely potential.

4. A pendulum having a time period 2 seconds is called a second’s pendulum.

Question 40. A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period T.

1. \(T \propto \sqrt{m}\)
2. \(T \propto \sqrt{\rho}\)
3. \(T \propto \frac{1}{\sqrt{A}}\)
4. \(T \propto \frac{1}{\sqrt{\rho}}\)

Answer:

1. \(T \propto \sqrt{m}\)

3. \(T \propto \frac{1}{\sqrt{A}}\)

4. \(T \propto \frac{1}{\sqrt{\rho}}\)

Question 41. A linear harmonic oscillator of force constant 2x 10⁶ N/m and amplitude 0.01m has a total mechanical energy of 160 J. Its

1. Maximum potential energy is 160J
2. Maximum kinetic energy is 160J
3. Maximum potential energy is 100J
4. Maximum potential energy is 0

Answer:

1. Maximum potential energy is 160J
2. Maximum kinetic energy is 160J

Question 42. A particle of mass m is attached to one end of a massless spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity u₀. When the speed of the particle is 0.5u₀, it collides elastically with a rigid wall. After this collision

1. The speed of the particle when it returns to its equilibrium position is u₀
2. The time at which the particle passes through the equilibrium position for the first time is t = \(pi \sqrt{\frac{m}{k}}\)
3. The time at which the maximum compression of the spring occurs is t = \(\frac{4 \pi}{3} \sqrt{\frac{m}{k}}\)
4. The time at which the particle passes through the equilibrium position for the second time is t = \(\frac{5 \pi}{3} \sqrt{\frac{m}{k}}\)

Answer:

1. The speed of the particle when it returns to its equilibrium position is u₀

4. The time at which the particle passes through the equilibrium position for the second time is t = \(\frac{5 \pi}{3} \sqrt{\frac{m}{k}}\)