JEE Physics Oscillations MCQs Set 22

Answer: (d) \( 2\pi\sqrt{\dfrac{m(YA+kL)}{YAk}} \)

Explanation:
1. Wire stiffness \( k_w=\dfrac{YA}L \), in series with the spring: \( \dfrac1{k_{eq}}=\dfrac1k+\dfrac L{YA}=\dfrac{YA+kL}{kYA} \).
2. \( T=2\pi\sqrt{\dfrac m{k_{eq}}}=2\pi\sqrt{\dfrac{m(YA+kL)}{YAk}} \).

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) \( 2\pi\sqrt{\dfrac M{\eta L}} \)

Explanation:
1. Shear: \( \eta=\dfrac{F/L^2}{x/L}\Rightarrow F=\eta L\,x \), so the effective spring constant is \( k=\eta L \).
2. \( T=2\pi\sqrt{\dfrac Mk}=2\pi\sqrt{\dfrac M{\eta L}} \).

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) \( \dfrac1{2\pi}\left(\dfrac{k+A\rho g}M\right)^{1/2} \)

Explanation:
1. For a displacement \( y \) the extra restoring force is the spring force \( ky \) plus the change in buoyancy \( A\rho g\,y \).
2. \( k_{eff}=k+A\rho g\Rightarrow f=\dfrac1{2\pi}\sqrt{\dfrac{k+A\rho g}M} \).

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: (c) \( \dfrac1{2\pi}\sqrt{\dfrac{3g}{2R}} \)

Explanation:
1. Mass \( M=\rho\cdot\tfrac12\cdot\tfrac43\pi R^3=\tfrac23\pi\rho R^3 \). For a small push \( y \) the water-plane area is \( \pi R^2 \), so the extra buoyancy is \( \rho g\pi R^2y \).
2. \( \omega^2=\dfrac{\rho g\pi R^2}{\tfrac23\pi\rho R^3}=\dfrac{3g}{2R}\Rightarrow f=\dfrac1{2\pi}\sqrt{\dfrac{3g}{2R}} \).

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) \( \sqrt{\dfrac{3g(d_2-d_1)}{2d_1L}} \)

Explanation:
1. Restoring torque about \( P \): \( (d_2-d_1)ALg\cdot\dfrac L2\sin\theta\approx\dfrac{(d_2-d_1)ALg\,L}2\theta \).
2. \( I=\dfrac13(d_1AL)L^2 \). \( \omega^2=\dfrac{(d_2-d_1)gL^2A L/2}{d_1AL^3/3}=\dfrac{3g(d_2-d_1)}{2d_1L} \).

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) 0.628 s

Explanation:
1. On the axis, \( F=-\dfrac{kQq\,x}{(R^2+x^2)^{3/2}}\approx-\dfrac{kQq}{R^3}x \) for \( x\ll R \).
2. \( \omega^2=\dfrac{kQq}{mR^3}=\dfrac{9\times10^9\times10^{-5}\times10^{-6}}{0.9\times10^{-3}\times1}=100\Rightarrow\omega=10 \) rad/s.
3. \( T=\dfrac{2\pi}{10}=0.628 s \).

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: (b) \( \dfrac1{2\pi}\sqrt{\dfrac{\gamma\left(p_0A^2+MgA\right)}{V_0M}} \)

Explanation:
1. Equilibrium gas pressure \( p=p_0+\dfrac{Mg}A \). Adiabatic: \( pV^\gamma=\) const, so \( dp=-\gamma p\dfrac{dV}V \).
2. For piston displacement \( x \): \( dV=Ax \), restoring force \( =A\,dp=\dfrac{\gamma pA^2}{V_0}x \).
3. \( \omega^2=\dfrac{\gamma pA^2}{MV_0}=\dfrac{\gamma(p_0A^2+MgA)}{MV_0} \).

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) \( \dfrac1{2\pi}\sqrt{\dfrac{YA}{mL}} \)

Explanation:
1. Wire acts as a spring with \( k=\dfrac{YA}L \).
2. \( f=\dfrac1{2\pi}\sqrt{\dfrac km}=\dfrac1{2\pi}\sqrt{\dfrac{YA}{mL}} \).

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) 4

Explanation:
1. \( \omega^2=\dfrac{YA}{mL}\Rightarrow Y=\dfrac{m\omega^2L}A \).
2. \( Y=\dfrac{0.1\times19600\times1}{4.9\times10^{-7}}=4\times10^9 \), so \( n = 4 \).

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) \( \dfrac{1}{n} \)

Explanation:
1. Force constant is inversely proportional to length: \( kl=\) const.
2. \( \dfrac{k_1}{k_2}=\dfrac{l_2}{l_1}=\dfrac1n \).

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) \( 10^{-5} \) K

Explanation:
1. Vibrational energy \( =\tfrac12kx^2=\tfrac12(800)(0.02)^2=0.16 J \), all converted to heat.
2. Heat capacity \( =1(4184)+0.5(400)=4384 J/K \). \( \Delta T=\dfrac{0.16}{4384}\approx3.6\times10^{-5} K \), of order \( 10^{-5} K \).

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: (c) \( \dfrac1{2\pi}\sqrt{\dfrac{6k}m} \)

Explanation:
1. For a small angle \( \theta \), each spring force is \( k\dfrac l2\theta \) acting at lever arm \( \dfrac l2 \): restoring torque \( =2k\dfrac{l^2}4\theta=\dfrac{kl^2}2\theta \).
2. \( I=\dfrac{ml^2}{12} \). \( \omega^2=\dfrac{kl^2/2}{ml^2/12}=\dfrac{6k}m \).

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: (d) \( \dfrac{F}{\sqrt{mk}} \)

Explanation:
1. The block moves to the maximum speed where \( kx=F \), i.e. \( x=\dfrac Fk \).
2. Work-energy: \( \tfrac12mv^2=Fx-\tfrac12kx^2=\dfrac{F^2}{2k}\Rightarrow v=\dfrac F{\sqrt{mk}} \).

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) \( \sqrt{50} \) m/s

Explanation:
1. Block position at \( t = 1 s \): \( x=4.9+0.2\cos\dfrac\pi3=5.0 m \) from \( O \).
2. Pebble must cover \( 10-5=5 m \) horizontally in \( 1 s \): \( v\cos45^\circ=5 \). It lands at the same level: \( v\sin45^\circ\cdot1-\tfrac12(10)(1)^2=0\Rightarrow v\sin45^\circ=5 \).
3. \( v=\sqrt{5^2+5^2}=\sqrt{50} m/s \).

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: (b) of the same frequency and with shifted mean position

Explanation:
1. The electric force QE is constant, so it does not change the force constant \( k \) or the mass, hence the frequency is unchanged.
2. The constant force shifts the equilibrium position by QE/k, so the mean position shifts.

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.