JEE Physics Oscillations MCQs Set 23

Answer: (c) \( \dfrac{k_2A}{k_1+k_2} \)

Explanation:
1. The two springs carry the same force: \( k_1x_1=k_2x_2 \) and \( x_1+x_2=A \).
2. The point \( P \) moves by \( x_1=\dfrac{k_2}{k_1+k_2}A \).

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.

Answer: (d) \( \dfrac1{2\pi}\sqrt{\dfrac{6k}M} \)

Explanation:
1. Each spring is stretched by \( \dfrac L2\theta \) and exerts torque \( k\dfrac{L^2}4\theta \); total restoring torque \( =\dfrac{kL^2}2\theta \).
2. \( \omega^2=\dfrac{kL^2/2}{ML^2/12}=\dfrac{6k}M \).

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.

Answer: (a) \( kA/2 \)

Explanation:
1. The two blocks move together with \( \omega^2=\dfrac k{2m} \); maximum acceleration \( =\omega^2A=\dfrac{kA}{2m} \).
2. Only friction accelerates the lower block \( P \): \( f_{max}=m\cdot\dfrac{kA}{2m}=\dfrac{kA}2 \).

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.

Answer: (b) \( \dfrac{3}{2} \) k

Explanation:
1. The long piece is \( \tfrac23 \) of the original length, and \( k\propto\dfrac1l \).
2. \( k'=\dfrac32k \).

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.

Answer: (c) \( \sqrt{k_2/k_1} \)

Explanation:
1. \( v_{max}=A\omega=A\sqrt{\dfrac km} \), equal for both.
2. \( \dfrac{A_M}{A_N}=\sqrt{\dfrac{k_2}{k_1}} \).

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.

Answer: (d) (a) and (d) only

Explanation:
1. Let \( \omega=\sqrt{k/m} \). The speed is \( 0.5u_0 \) when \( \omega t=\dfrac\pi3 \); the wall reverses the velocity and no energy is lost, so the speed at equilibrium is again \( u_0 \): \( (a) \) true.
2. After the bounce it takes another \( \dfrac\pi{3\omega} \) to reach equilibrium, so the first pass is at \( t=\dfrac{2\pi}{3}\sqrt{\dfrac mk} \): \( (b) \) false. The second pass is half a period later: \( \dfrac{2\pi}3+\pi=\dfrac{5\pi}3\sqrt{\dfrac mk} \): \( (d) \) true.
3. Maximum compression occurs at the wall at \( t=\dfrac\pi{3\omega} \), not at \( \dfrac{4\pi}3 \): \( (c) \) false.

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.

Answer: (a) 2.09

Explanation:
1. Elastic collision: \( v_2=\dfrac{2(1)}{3}(2)=\dfrac43 m/s \) and \( v_1=\dfrac{1-2}{3}(2)=-\dfrac23 m/s \) (the 1 kg block rebounds).
2. The 2 kg block does SHM with \( \omega=\sqrt{\dfrac22}=1 \) rad/s and returns to the unstretched position after half a period, \( t=\pi s \).
3. At that time the 1 kg block is \( \dfrac23\pi=2.09 m \) away.

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.

Answer: (b) \( \dfrac g{\omega^2} \)

Explanation:
1. Speed at \( y \): \( v^2=\omega^2(A^2-y^2) \). Height reached above the mean position \( H=y+\dfrac{v^2}{2g}=y+\dfrac{\omega^2(A^2-y^2)}{2g} \).
2. \( \dfrac{dH}{dy}=1-\dfrac{\omega^2y}g=0\Rightarrow y=\dfrac g{\omega^2} \).

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.

Answer: (c) \( \dfrac1\pi \) Hz

Explanation:
1. If a ball moves a distance \( s \) along the pipe, each spring changes its length by \( 2s \) (both balls move), so each spring changes force by 2ks.
2. Ball \( B \) feels two such forces in the same sense: \( F=4ks \), so \( \omega^2=\dfrac{4k}m=\dfrac{4(0.1)}{0.1}=4\Rightarrow\omega=2 \) rad/s.
3. \( f=\dfrac\omega{2\pi}=\dfrac1\pi \) Hz.

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.

Answer: (d) 0.0628 m/s

Explanation:
1. The amplitude of each ball along the pipe is \( s_0=R\theta=0.06\times\dfrac\pi6=0.0314 m \).
2. At the equilibrium (ends of the diameter) the speed is maximum: \( v=\omega s_0=2\times0.0314=0.0628 m/s \).

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.

Answer: (a) \( 3.9\times10^{-4} \) J

Explanation:
1. At the equilibrium position all the energy is kinetic: each ball moves at \( 0.0628 m/s \).
2. \( E=2\times\tfrac12(0.1)(0.0628)^2=3.9\times10^{-4} J \).

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.

Answer: (b) 2.83 s

Explanation:
1. Free motion between the springs: \( \dfrac{0.6}{1.2}=0.5 s \) each way, i.e. \( 1.0 s \) for the round trip.
2. Half a period in each spring: \( \pi\sqrt{\dfrac{0.2}{1.8}}=\dfrac\pi3=1.047 s \) and \( \pi\sqrt{\dfrac{0.2}{3.2}}=\dfrac\pi4=0.785 s \).
3. \( T=1.0+1.047+0.785\approx2.83 s \).

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.

Answer: (c) \( \omega=\sqrt{\dfrac k{m_2}},\ A=\dfrac{m_1g}k \)

Explanation:
1. After \( m_1 \) is removed the oscillating mass is \( m_2 \), so \( \omega=\sqrt{k/m_2} \).
2. Initial stretch \( \dfrac{(m_1+m_2)g}k \); new equilibrium stretch \( \dfrac{m_2g}k \). Amplitude \( =\dfrac{m_1g}k \).

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.

Answer: (d) 1.6 kg

Explanation:
1. \( T\propto\sqrt M \): \( \left(\dfrac32\right)^2=\dfrac{M+2}M \).
2. \( 2.25M=M+2\Rightarrow M=1.6 \) kg.

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.

Answer: (a) \( 4T\sqrt{\dfrac1{15}} \)

Explanation:
1. Effective gravity \( g'=g\left(1-\dfrac{\rho_{liquid}}{\rho_{bob}}\right)=\dfrac{15}{16}g \).
2. \( T'=T\sqrt{\dfrac g{g'}}=T\sqrt{\dfrac{16}{15}}=4T\sqrt{\dfrac1{15}} \).

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.