JEE Physics Oscillations MCQs Set 13

Answer: (a) the particle is at \( x=\frac a2 \) and moving in the +X direction

Explanation:
1. For \( x=a\sin\phi \) the phase is \( \phi=\frac\pi6 \), so \( x=a\sin30^\circ=\frac a2 \).
2. Velocity \( v=a\omega\cos\phi=a\omega\cos30^\circ>0 \), so the particle moves in the +X direction.

Teacher's Note:
1. The sign of \( \cos\phi \) gives the direction of velocity, and \( \sin\phi \) gives the position.

Answer: (a) \( \frac\pi2\ \text{s} \)

Explanation:
1. The ellipse crosses the x-axis at \( x=\pm A \), so \( A=0.1 \) m, and the v-axis at \( v=\pm v_{max} \), so \( v_{max}=0.4 \) m/s.
2. \( \omega=\frac{v_{max}}{A}=\frac{0.4}{0.1}=4 \) rad/s.
3. \( T=\frac{2\pi}{\omega}=\frac\pi2 \) s.

Teacher's Note:
1. From \( v^2=\omega^2(A^2-x^2) \) the v-x graph is an ellipse with semi-axes A and \( A\omega \). Convert cm to m first.

Answer: (a) \( v \)

Explanation:
1. The maximum velocity is \( A\omega \), which depends only on the amplitude and the angular frequency.
2. Both pendulums are identical with the same amplitude, so \( A\omega \) is the same.
3. Phase difference does not change the maximum value; the other maximum velocity is \( v \).

Teacher's Note:
1. Phase difference shifts when the maximum occurs, not how large it is.

Answer: (a) a straight line through the origin with negative slope

Explanation:
1. \( a=-\omega^2x \), a linear relation with negative slope that passes through the origin.
2. So the graph of a against x is a straight line through the origin with negative slope.

Teacher's Note:
1. Acceleration is proportional to displacement and opposite to it; the negative slope is the restoring nature of the force.

Answer: (d) \( \frac{2\pi\sqrt2}{5}\ \text{s} \)

Explanation:
1. \( U=\frac12kx^2 \) gives \( 1.0=\frac12k(0.2)^2 \), so \( k=50 \) N/m.
2. \( T=2\pi\sqrt{\frac mk}=2\pi\sqrt{\frac4{50}}=2\pi\frac{\sqrt2}5 \).
3. \( T=\frac{2\pi\sqrt2}{5} \) s.

Teacher's Note:
1. Find the force constant from the energy first, then use the spring formula for the period.

Answer: (d) 72 s

Explanation:
1. \( T\propto\sqrt L \). New length is \( 1.44L \), so \( T'=\sqrt{1.44}\,T=1.2T \).
2. \( T'=1.2\times60 \) s.
3. \( T'=72 \) s.

Teacher's Note:
1. A 44 percent increase in length gives a 20 percent increase in period.

Answer: (c) \( \sqrt3:2 \)

Explanation:
1. Arrangement 1: two springs in series give \( K_1=\frac K2 \).
2. Arrangement 2: the parallel pair gives \( 2K \), in series with K gives \( K_2=\frac{2K\cdot K}{3K}=\frac{2K}3 \).
3. \( f\propto\sqrt K \): \( \frac{f_1}{f_2}=\sqrt{\frac{K/2}{2K/3}}=\sqrt{\frac34}=\frac{\sqrt3}2 \).

Teacher's Note:
1. Reduce each spring network to one equivalent constant, then use \( f\propto\sqrt K \).

Answer: (d) \( \frac{K_1K_2}{K_1+K_2} \)

Explanation:
1. Under the same displacement x, each spring has the same tension F in series; extensions are \( \frac F{K_1} \) and \( \frac F{K_2} \).
2. Total extension \( \frac F{K_{eff}}=\frac F{K_1}+\frac F{K_2} \).
3. \( K_{eff}=\frac{K_1K_2}{K_1+K_2} \).

Teacher's Note:
1. Springs in series combine like resistors in parallel.

Answer: (b) the effective value of g becomes zero in the artificial satellite

Explanation:
1. A pendulum works because gravity gives a restoring torque.
2. In a freely orbiting satellite the effective g is zero, so there is no restoring force and the pendulum does not oscillate.

Teacher's Note:
1. Orbiting bodies are in free fall, which is why they feel weightless.

Answer: (a) i, iii, iv

Explanation:
1. Free vibrations of an ideal oscillator have constant amplitude; maintained vibrations are supplied with energy to offset losses; forced vibrations have constant amplitude set by the driving force.
2. Damped vibrations lose energy, so their amplitude decreases.

Teacher's Note:
1. Only damped motion has a decaying amplitude.

Answer: (d) \( T_{st}=T_{ma}=T_{sp}=T_{is} \)

Explanation:
1. A satellite just above the surface: \( T=2\pi\sqrt{R/g} \). Oscillation of a ball in a tunnel through the centre also has \( T=2\pi\sqrt{R/g} \).
2. A pendulum of length R in a uniform field g gives \( T=2\pi\sqrt{R/g} \).
3. An infinitely long pendulum in the earth's field has \( T=2\pi\sqrt{R/g} \) (from \( \frac1T\propto\sqrt{\frac1L+\frac1R} \) with \( L\to\infty \)). All four are equal, about 84.6 minutes.

Teacher's Note:
1. All four give the same period \( 2\pi\sqrt{R/g} \).