JEE Physics Oscillations MCQs Set 28

Practice MCQs for JEE Physics Oscillations

Access targeted multiple-choice questions for Oscillations designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Oscillations Questions and Solutions

Access the complete set of multiple-choice questions for Oscillations below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: A mass M attached to a spring oscillates with a period of 2 s. When the mass is increased by 4 kg, the period increases by 1 s. Assuming Hooke's law holds, the initial mass M was
(a) 3.2 kg
(b) 1 kg
(c) 2 kg
(d) 8 kg
Show Answer & Explanation

Answer: (a) 3.2 kg

Explanation:
1. \( T=2\pi\sqrt{M/k} \), so \( T^2\propto M \).
2. \( (3/2)^2=(M+4)/M \), so \( 2.25M=M+4 \).
3. \( M=4/1.25=3.2 \) kg.

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Question 2: The period of a mass m suspended by a spring is T. If the spring is cut to half its length and made to oscillate with double the mass suspended, the new period is
(a) 8T
(b) T
(c) 4T
(d) T/2
Show Answer & Explanation

Answer: (b) T

Explanation:
1. Cutting the spring in half doubles its force constant: \( k'=2k \).
2. \( T'=2\pi\sqrt{2m/2k}=2\pi\sqrt{m/k}=T \).

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Question 3: Two blocks of masses \( m_1=1 \) kg and \( m_2=2 \) kg are joined by a spring of constant 24 N/m on a frictionless horizontal surface. Block \( m_1 \) is given an initial velocity of 12 cm/s towards the right. The amplitude of oscillation is
(a) 2 cm
(b) 1 cm
(c) 3 cm
(d) 4 cm
Show Answer & Explanation

Answer: (a) 2 cm

Explanation:
1. Reduced mass \( \mu=\dfrac{1\times2}{3}=\dfrac23 \) kg. In the centre of mass frame the relative speed is 0.12 m/s.
2. Energy of oscillation \( \tfrac12\mu v_{rel}^2=\tfrac12kA^2 \).
3. \( A=v_{rel}\sqrt{\mu/k}=0.12\sqrt{(2/3)/24}=0.12\times\tfrac16=0.02 \) m = 2 cm.

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Question 4: A mass of 1 kg suspended from a spring of force constant 400 N/m executes simple harmonic oscillation. When the total energy of the oscillator is 2 J, the maximum acceleration of the mass is
(a) 2 m/s\(^2\)
(b) 4 m/s\(^2\)
(c) 40 m/s\(^2\)
(d) 400 m/s\(^2\)
Show Answer & Explanation

Answer: (c) 40 m/s\(^2\)

Explanation:
1. \( E=\tfrac12kA^2 \), so \( A=\sqrt{2E/k}=\sqrt{4/400}=0.1 \) m.
2. \( a_{max}=\omega^2A=(k/m)A=400\times0.1=40 \) m/s\(^2\).

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Question 5: A uniform spring of force constant k is cut into two pieces whose lengths are in the ratio 1 : 2. The ratio of the force constants of the shorter and longer pieces is
(a) 2 : 1
(b) 1 : 2
(c) 1 : 3
(d) 2 : 3
Show Answer & Explanation

Answer: (a) 2 : 1

Explanation:
1. The force constant is inversely proportional to length.
2. Shorter piece has length \( l/3 \), so \( k_1=3k \). Longer piece has length \( 2l/3 \), so \( k_2=3k/2 \).
3. \( k_1:k_2=2:1 \).

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Question 6: One end of a spring of force constant k is fixed to a vertical wall and the other end to a block of mass m on a smooth horizontal surface. Another wall is at a distance \( x_0 \) from the block's equilibrium position. The spring is compressed by \( 2x_0 \) and released. The time taken to strike the wall is
(a) \( \dfrac{2\pi}{3}\sqrt{\dfrac mk} \)
(b) \( \dfrac16\pi\sqrt{\dfrac km} \)
(c) \( \sqrt{\dfrac mk} \)
(d) \( \dfrac\pi4\sqrt{\dfrac km} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{2\pi}{3}\sqrt{\dfrac mk} \)

Explanation:
1. The amplitude is \( A=2x_0 \). The block starts at \( x=-A \) and reaches the wall at \( x=+x_0=+A/2 \).
2. Time from \( -A \) to 0 is \( T/4 \); from 0 to \( A/2 \) is \( T/12 \).
3. Total \( t=T/3=\dfrac{2\pi}{3}\sqrt{m/k} \).

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Question 7: A simple pendulum of length l and bob mass m makes an angle \( \theta \) with the vertical. The restoring force acting on the bob is
(a) \( -mg\sin\theta \)
(b) \( mg\tan\theta \)
(c) \( mg\sin\theta \)
(d) \( -mg\cos\theta \)
Show Answer & Explanation

Answer: (a) \( -mg\sin\theta \)

Explanation:
1. The tangential component of the weight along the arc is \( mg\sin\theta \), directed towards the mean position.
2. With the displacement-based sign convention the restoring force is \( -mg\sin\theta \).

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Question 8: A man measures the period of a simple pendulum inside a stationary lift and finds it to be T. If the lift accelerates upwards with acceleration g/4, the new period will be
(a) \( \dfrac{2T}{\sqrt5} \)
(b) \( 2T\sqrt5 \)
(c) T
(d) \( \dfrac T4 \)
Show Answer & Explanation

Answer: (a) \( \dfrac{2T}{\sqrt5} \)

Explanation:
1. Effective gravity \( g'=g+g/4=5g/4 \).
2. \( T'=2\pi\sqrt{l/g'}=T\sqrt{g/g'}=T\sqrt{4/5}=\dfrac{2T}{\sqrt5} \).

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Question 9: If the length of a second's pendulum is increased by 2%, the number of seconds it loses per day is about
(a) 864 s
(b) 3927 s
(c) 3427 s
(d) 3737 s
Show Answer & Explanation

Answer: (a) 864 s

Explanation:
1. \( T\propto\sqrt l \), so a 2% increase in length raises T by about 1%.
2. Clock loses \( 0.01\times86400=864 \) s per day.

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Question 10: A pendulum bob of mass m hangs from a fixed point by a light thread of length l. A horizontal speed \( v_0 \) is given so that the bob just reaches the horizontal position (string taut). Then \( v_0 \) equals
(a) \( \sqrt{2gl} \)
(b) \( mgl \)
(c) \( \sqrt{gl} \)
(d) \( 2gl \)
Show Answer & Explanation

Answer: (a) \( \sqrt{2gl} \)

Explanation:
1. Energy conservation: \( \tfrac12mv_0^2=mgl \).
2. \( v_0=\sqrt{2gl} \).

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Question 11: The bob of a pendulum of length l is pulled aside through an angle \( \theta \) and released. The bob passes through the equilibrium position with speed v equal to
(a) \( \sqrt{2gl(1-\cos\theta)} \)
(b) \( \sqrt{2gl(1+\sin\theta)} \)
(c) \( \sqrt{2gl(1-\sin\theta)} \)
(d) \( \sqrt{2gl(1+\cos\theta)} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{2gl(1-\cos\theta)} \)

Explanation:
1. The height gained from the lowest point is \( h=l(1-\cos\theta) \).
2. \( \tfrac12mv^2=mgl(1-\cos\theta) \), so \( v=\sqrt{2gl(1-\cos\theta)} \).

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Question 12: In damped oscillation the amplitude of oscillations is reduced to one-third of its initial value \( a_0 \) at the end of 100 oscillations. When the oscillator completes 200 oscillations its amplitude must be
(a) \( \dfrac{a_0}{2} \)
(b) \( \dfrac{a_0}{4} \)
(c) \( \dfrac{a_0}{9} \)
(d) \( \dfrac{a_0}{6} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{a_0}{9} \)

Explanation:
1. The amplitude decays exponentially: \( a=a_0e^{-bt} \).
2. After 100 oscillations the factor is 1/3; after 200 oscillations the same factor applies again: \( (1/3)^2=1/9 \).

Teacher's Note:
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Question 13: If a simple pendulum is taken to a place where g decreases by 2%, the time period
(a) increases by 1%
(b) increases by 0.5%
(c) increases by 2.0%
(d) decreases by 0.5%
Show Answer & Explanation

Answer: (a) increases by 1%

Explanation:
1. \( T\propto g^{-1/2} \), so \( \Delta T/T=-\tfrac12\Delta g/g=+1\% \).

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Question 14: A pendulum clock that keeps good time on the earth is placed on the moon, where an object weighs only one-sixth as much as on the earth. How many clock seconds does it tick out in an actual time of 1 minute?
(a) 24.5
(b) 12.25
(c) 2.45
(d) 0.245
Show Answer & Explanation

Answer: (a) 24.5

Explanation:
1. \( T\propto1/\sqrt g \), so the period on the moon is \( \sqrt6 \) times longer.
2. Number of ticks in 60 s \( =60/\sqrt6=24.5 \).

Teacher's Note:
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Question 15: A heavy sphere of mass m is suspended by a string of length l and made to revolve in a horizontal circle with the string inclined at an angle \( \theta \) to the vertical. Its period of revolution is
(a) \( 2\pi\sqrt{\dfrac lg} \)
(b) \( 2\pi\sqrt{\dfrac{l\cos\theta}{g}} \)
(c) \( 2\pi\sqrt{\dfrac{l\sin\theta}{g}} \)
(d) \( 2\pi\sqrt{\dfrac{l\tan\theta}{g}} \)
Show Answer & Explanation

Answer: (b) \( 2\pi\sqrt{\dfrac{l\cos\theta}{g}} \)

Explanation:
1. \( T\cos\theta=mg \) and \( T\sin\theta=m\omega^2l\sin\theta \), so \( T=m\omega^2l \).
2. \( \omega^2=g/(l\cos\theta) \), so the period is \( 2\pi\sqrt{l\cos\theta/g} \).

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Question 16: The instantaneous displacement of a simple harmonic oscillator is \( x=A\cos(\omega t+\pi/4) \). Its speed will be maximum at time
(a) \( \dfrac{\pi}{4\omega} \)
(b) \( \dfrac{\pi}{2\omega} \)
(c) \( \dfrac{\pi}{\omega} \)
(d) \( \dfrac{2\pi}{\omega} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{\pi}{4\omega} \)

Explanation:
1. The speed is maximum at the mean position, where \( \cos(\omega t+\pi/4)=0 \).
2. \( \omega t+\pi/4=\pi/2 \), so \( t=\pi/(4\omega) \).

Teacher's Note:
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Question 17: The period of oscillation of a mass m suspended from a spring is 2 s. If another mass of 2 kg is suspended along with it, the period increases by 1 s. The mass m is
(a) 1.6 kg
(b) 2 kg
(c) 1 kg
(d) 2.6 kg
Show Answer & Explanation

Answer: (a) 1.6 kg

Explanation:
1. \( T^2\propto m \): \( (3/2)^2=(m+2)/m \).
2. \( 2.25m=m+2 \), so \( m=1.6 \) kg.

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Question 18: Two pendulums begin to swing simultaneously. The first makes 9 full oscillations while the other makes 7. The ratio of the lengths of the two pendulums is
(a) \( \dfrac97 \)
(b) \( \dfrac{49}{81} \)
(c) \( \dfrac79 \)
(d) \( \dfrac{81}{49} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{49}{81} \)

Explanation:
1. In the same time, \( n_1T_1=n_2T_2 \), so \( T_1/T_2=7/9 \).
2. \( L\propto T^2 \), so \( L_1/L_2=49/81 \).

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Question 19: A bottle weighing 220 g, with area of cross-section 50 cm\(^2\) and height 4 cm, oscillates vertically on the surface of water. Its frequency of oscillation is about (density of water 1000 kg/m\(^3\), g = 9.8 m/s\(^2\))
(a) 1.5 Hz
(b) 3.5 Hz
(c) 2.5 Hz
(d) 4.5 Hz
Show Answer & Explanation

Answer: (c) 2.5 Hz

Explanation:
1. The restoring force constant is \( k=\rho gA=1000\times9.8\times50\times10^{-4}=49 \) N/m.
2. \( f=\dfrac1{2\pi}\sqrt{k/m}=\dfrac1{2\pi}\sqrt{49/0.22}\approx2.4 \) Hz, closest to 2.5 Hz.

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Question 20: A body of mass 4.9 kg hangs from a spring and oscillates with a period of 0.5 s. When the body is removed, the spring shortens by (take g = 10 m/s\(^2\), \( \pi^2=10 \))
(a) 6.3 m
(b) 0.63 m
(c) 63.5 cm
(d) 6.25 cm
Show Answer & Explanation

Answer: (d) 6.25 cm

Explanation:
1. \( k=4\pi^2m/T^2=40\times4.9/0.25=784 \) N/m.
2. Extension under the weight \( =mg/k=49/784=0.0625 \) m = 6.25 cm.

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