JEE Physics Oscillations MCQs Set 21

Answer: (a) \( K_2=\dfrac{K_1}{2} \)

Explanation:
1. \( K_{max}=\tfrac12m\omega^2a^2 \) with \( \omega^2=\dfrac gl \) and the same linear amplitude a.
2. \( K_{max}\propto\dfrac1l \): doubling \( l \) halves the maximum kinetic energy.

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (b) \( \dfrac{1}{3} \)

Explanation:
1. At \( t=210 s \), \( \dfrac{\pi t}{90}=\dfrac{7\pi}3 \), equivalent to \( \dfrac\pi3 \).
2. \( U=\tfrac12kA^2\sin^2\tfrac\pi3 \), \( K=\tfrac12kA^2\cos^2\tfrac\pi3 \), so \( \dfrac KU=\cot^2\dfrac\pi3=\dfrac13 \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (c) \( \dfrac{A}{\sqrt2} \)

Explanation:
1. \( \tfrac12kx^2=\tfrac12k(A^2-x^2)\Rightarrow x^2=\dfrac{A^2}2 \).
2. \( x=\pm\dfrac A{\sqrt2} \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (d) PE is an upward-opening parabola with zero at d = 0, KE is a downward-opening parabola with its maximum at d = 0, and they are equal at \( d=\pm A/\sqrt2 \)

Explanation:
1. \( PE=\tfrac12m\omega^2d^2 \) and \( KE=\tfrac12m\omega^2(A^2-d^2) \), both parabolas in \( d \).
2. At \( d = 0 \), PE = 0 and KE is maximum; at \( d=\pm A \), KE = 0 and PE is maximum; they are equal at \( d=\pm\dfrac A{\sqrt2} \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (a) 2f

Explanation:
1. \( KE=\tfrac12m\omega^2A^2\cos^2\omega t=\tfrac14m\omega^2A^2(1+\cos2\omega t) \).
2. The kinetic energy therefore varies with angular frequency \( 2\omega \), i.e. at frequency \( 2f \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (b) (b) and (c) only

Explanation:
1. Energy of oscillation \( =\tfrac12kA^2=\tfrac12(2\times10^6)(10^{-2})^2=100 J \), which is the maximum kinetic energy: \( (b) \) true.
2. At the mean position \( U=160-100=60 J \) (minimum). At the extremes \( U=160 J \) (maximum): \( (c) \) true.

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (c) 0.06

Explanation:
1. \( \omega=2\pi f=50 \) rad/s. Total energy \( E=0.5+0.4=0.9 J \).
2. \( \tfrac12m\omega^2A^2=0.9\Rightarrow A^2=\dfrac{0.9}{0.1\times2500}=0.0036\Rightarrow A=0.06 m \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (d) KE is maximum at t = 0, falls to zero at t = T/4 and is again maximum at t = T/2, so it repeats with period T/2

Explanation:
1. Starting at equilibrium the speed is maximum, so \( KE=KE_{max}\cos^2\dfrac{2\pi t}T \).
2. It is zero at \( t=\dfrac T4 \) (extreme position) and returns to maximum at \( t=\dfrac T2 \), so its period is \( T/2 \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (a) \( -\dfrac{\sqrt3}{32}\pi^2 \) cm s\(^{-2}\)

Explanation:
1. \( x=\sin\dfrac{2\pi t}8=\sin\dfrac{\pi t}4 \) cm, so \( \omega=\dfrac\pi4 \).
2. \( a=-\omega^2x=-\dfrac{\pi^2}{16}\sin\dfrac\pi3=-\dfrac{\pi^2}{16}\cdot\dfrac{\sqrt3}2=-\dfrac{\sqrt3\pi^2}{32} \) cm \( s ^{-2} \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (b) 7 s

Explanation:
1. Amplitude \( =e^{-0.1t}=\tfrac12 \).
2. \( t=\dfrac{\ln2}{0.1}=6.9\approx7 s \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (c) 10 cm, \( \dfrac23 \) s

Explanation:
1. \( \sin\theta+\sqrt3\cos\theta=2\sin\left(\theta+\dfrac\pi3\right) \), so \( y=10\sin\left(3\pi t+\dfrac\pi3\right) \).
2. Amplitude 10 cm; \( T=\dfrac{2\pi}{3\pi}=\dfrac23 s \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (d) \( \dfrac{8\pi}{3} \)

Explanation:
1. Speed equals acceleration: \( \omega\sqrt{A^2-x^2}=\omega^2x \). In SI: \( \omega\sqrt{0.05^2-0.04^2}=\omega^2(0.04)\Rightarrow\omega=\dfrac{0.03}{0.04}=\dfrac34 \) rad/s.
2. \( T=\dfrac{2\pi}\omega=\dfrac{8\pi}3 s \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (a) \( 92\pm2 \) s

Explanation:
1. Mean \( =\dfrac{90+91+92+95}4=92 s \).
2. Deviations: 2, 1, 0, 3 give a mean absolute error of \( 1.5 s \), which exceeds the least count \( (1 s) \). Reported to the precision of the clock (whole seconds), it is \( 92\pm2 s \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (b) \( \dfrac{6}{5} \)

Explanation:
1. Acceleration of the support \( =2k=2 m/s^{2} \) upward, so \( g_{eff}=10+2=12 m/s^{2} \).
2. \( T^2\propto\dfrac1g\Rightarrow\dfrac{T_1^2}{T_2^2}=\dfrac{g_{eff}}g=\dfrac{12}{10}=\dfrac65 \).

Teacher's Note:
1. In SHM total energy is constant: KE = \( (1/2)k(A^2 - x^2) \) and PE = \( (1/2)kx^2 \).
2. Use \( x = A \sin(\omega t) \) or \( A \cos(\omega t) \) according to the starting position, and substitute the given time or displacement.

Answer: (c) \( 2\pi\sqrt{\dfrac L{g\cos\alpha}} \)

Explanation:
1. The vehicle accelerates down the slope with \( g\sin\alpha \). In its frame the pseudo-force cancels the component of gravity along the slope.
2. \( g_{eff}=g\cos\alpha \) (perpendicular to the incline), so \( T=2\pi\sqrt{\dfrac L{g\cos\alpha}} \).

Teacher's Note:
1. Identify the restoring force per unit displacement (the effective force constant) and the inertia, then use \( \omega = \sqrt{k_{eff}/m_{eff}} \).
2. Check the answer by testing a limiting case or verifying the dimensions of the result.