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
(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.
Teacher's Note:
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110. .
(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. \( T'=2\pi\sqrt{2m/2k}=2\pi\sqrt{m/k}=T \).
Teacher's Note:
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105. .
(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.
Teacher's Note:
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135. .
(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\(^2\).
Teacher's Note:
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(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. \( k_1:k_2=2:1 \).
Teacher's Note:
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115. .
(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 \( t=T/3=\dfrac{2\pi}{3}\sqrt{m/k} \).
Teacher's Note:
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128. .
(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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120. .
(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. \( T'=2\pi\sqrt{l/g'}=T\sqrt{g/g'}=T\sqrt{4/5}=\dfrac{2T}{\sqrt5} \).
Teacher's Note:
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122. .
(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.
Teacher's Note:
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152. .
(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. \( v_0=\sqrt{2gl} \).
Teacher's Note:
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89. .
(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 \( v=\sqrt{2gl(1-\cos\theta)} \).
Teacher's Note:
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112. .
(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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84. .
(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 \( \Delta T/T=-\tfrac12\Delta g/g=+1\% \).
Teacher's Note:
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3. f
4. a
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106. .
(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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86. .
(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 \( 2\pi\sqrt{l\cos\theta/g} \).
Teacher's Note:
1. T
2. h
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102. .
(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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106. .
(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.
Teacher's Note:
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96. .
(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 \( L_1/L_2=49/81 \).
Teacher's Note:
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119. .
(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.
Teacher's Note:
1. A
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106. .
(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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