JEE Physics Oscillations MCQs Set 10

Answer: (b) \( \frac{mg}{k}\sqrt{1+\frac{2hk}{mg}} \)

Explanation:
1. The ball hits the pan with speed \( v=\sqrt{2gh} \). The equilibrium position with the ball on the pan is a distance \( x_0=\frac{mg}{k} \) below the pan's original position.
2. Energy relative to the new equilibrium: \( \tfrac12kA^2=\tfrac12mv^2+\tfrac12kx_0^2 \).
3. \( A^2=\frac{2mgh}{k}+\frac{m^2g^2}{k^2}=\left(\frac{mg}{k}\right)^2\left(1+\frac{2hk}{mg}\right) \).
4. \( A=\frac{mg}{k}\sqrt{1+\frac{2hk}{mg}} \).

Teacher's Note:
1. Because the pan is massless, the ball keeps its speed when it lands; it sits at distance mg/k from the new mean position.
2. Use energy conservation measured from the equilibrium position of the loaded spring.

Answer: (a) an ellipse

Explanation:
1. The two motions have the same frequency and equal amplitude with phase difference \( \pi/4 \), which is neither 0, \( \pi \), nor \( \pi/2 \).
2. Eliminating t gives \( x^2+y^2-\sqrt2\,xy=2 \), the equation of a tilted ellipse.

Teacher's Note:
1. Phase difference 0 or pi gives a straight line, pi/2 with equal amplitudes gives a circle, anything else gives an ellipse.
2. Equal frequencies are needed for a closed (stable) path.

Answer: (d) \( \sqrt3 \)

Explanation:
1. \( A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}=\sqrt{1+1+2\cos60^\circ} \).
2. \( =\sqrt{3} \).

Teacher's Note:
1. Superposition of two SHMs of the same frequency gives another SHM with amplitude given by the phasor addition formula.
2. Here the phase difference is 60 degrees.

Answer: (d) \( m\omega^2A^2 \)

Explanation:
1. The phase difference is \( \pi/2 \), so the resultant amplitude is \( \sqrt{A^2+A^2}=\sqrt2A \).
2. \( E=\tfrac12m\omega^2(\sqrt2A)^2=m\omega^2A^2 \).

Teacher's Note:
1. Energy of SHM is (1/2) m w^2 times the square of the resultant amplitude.
2. Perpendicular-phase components add in quadrature.

Answer: (c) 8 unit

Explanation:
1. Energy \( E=\tfrac12m\omega^2A^2 \) and the particles are identical, so equal energy needs \( \omega_1A_1=\omega_2A_2 \).
2. \( 10\times4=\omega\times5\Rightarrow\omega=8 \).

Teacher's Note:
1. For identical masses the energy depends only on the product of angular frequency and amplitude (the maximum speed).
2. The initial phase does not affect the energy.

Answer: (b) the frequency is doubled while the amplitude is halved

Explanation:
1. \( v_{max}=\omega A \) constant and \( a_{max}=\omega^2A=\omega\,(\omega A) \).
2. Doubling \( a_{max} \) with constant \( \omega A \) needs \( \omega \) doubled.
3. Then \( A \) must be halved to keep \( \omega A \) fixed.

Teacher's Note:
1. Write a_max as w times v_max: with v_max fixed, a_max is directly proportional to w.
2. Check the product w A after the change.

Answer: (a) 5A

Explanation:
1. The motion is SHM about \( x=A \) with amplitude A, starting at the mean position (\( x=A \) at t = 0).
2. The time \( \frac{2.5\pi}{\omega}=1.25\,T \) with \( T=\frac{2\pi}{\omega} \).
3. One full period covers 4A and a further quarter period (mean to extreme) covers A: total \( 5A \).

Teacher's Note:
1. Identify the mean position from the constant term: it is x = A.
2. Distance in one quarter period starting from the mean position is exactly A.

Answer: (a) \( v_{max} \)

Explanation:
1. Velocity \( v=0 \) is acquired twice per cycle (at both extremes), so the smallest gap is T/2.
2. A velocity between 0 and \( \pm v_{max} \) is obtained at two points per cycle with unequal gaps, the smaller being less than T.
3. Only the maximum velocity \( v_{max} \) (at the mean position, moving in one direction) occurs once per cycle, so its repeat interval is exactly T.

Teacher's Note:
1. A velocity value is repeated twice per cycle unless it is the maximum (or minimum) velocity.
2. Consider the velocity-time graph, a cosine curve with a single maximum per cycle.

Answer: (d) its amplitude will change and become \( \sqrt2 \) times its previous amplitude

Explanation:
1. At an extreme the energy is all potential: \( E=\tfrac12kA^2 \). The energy at the mean position is the same total energy E.
2. Adding E gives a new total \( 2E=\tfrac12kA'^2 \), so \( A'=\sqrt2A \).
3. The period \( 2\pi\sqrt{m/k} \) is independent of amplitude and stays 2 s.

Teacher's Note:
1. Total energy of SHM is proportional to the square of amplitude.
2. Period of SHM does not depend on energy or amplitude.

Answer: (c) the magnitude of the acceleration at a distance 3 cm from the fixed point is 0.27 m/s²

Explanation:
1. Differentiating \( v^2=108-9x^2 \) with respect to x: \( 2v\frac{dv}{dx}=-18x \) so \( a=v\frac{dv}{dx}=-9x \).
2. So the motion is SHM about the fixed point x = 0 with \( \omega^2=9 \) and amplitude \( A=\sqrt{12}\ \text{m} \) (not 4 cm).
3. At 3 cm: \( |a|=9\times0.03=0.27\ \text{m/s}^2 \).

Teacher's Note:
1. From v^2 as a function of x, a = v dv/dx = (1/2) d(v^2)/dx.
2. Compare with v^2 = w^2 (A^2 - x^2) to read w and A.

Answer: (b) \( a/\sqrt2 \)

Explanation:
1. \( x=a\sin\omega t=a\sin\left(\frac{2\pi}{T}\cdot\frac T8\right)=a\sin45^\circ \).
2. \( x=\frac{a}{\sqrt2} \).

Teacher's Note:
1. Time measured from the mean position means x = a sin(wt).
2. T/8 corresponds to a phase of 45 degrees.

Answer: (b) \( \frac{4\sqrt2A}{T} \)

Explanation:
1. For maximum displacement in T/4 the motion should be symmetric about the mean position: the phase covered is \( 90^\circ \), from \( -45^\circ \) to \( +45^\circ \).
2. Displacement \( =2A\sin45^\circ=\sqrt2A \).
3. \( v_{avg,max}=\frac{\sqrt2A}{T/4}=\frac{4\sqrt2A}{T} \).

Teacher's Note:
1. The particle is fastest near the mean position, so the largest displacement in a given time is found by centring the interval on the mean position.
2. Average velocity here is displacement divided by time.

Answer: (a) \( \sqrt2+1 \)

Explanation:
1. With \( x=A\sin\frac{2\pi}{8}t=A\sin\frac{\pi t}{4} \):
2. at t = 1 s \( x=A\sin45^\circ=\frac{A}{\sqrt2} \); at t = 2 s \( x=A \).
3. Distance in 1st second: \( \frac{A}{\sqrt2} \); in 2nd second: \( A-\frac A{\sqrt2} \).
4. \( \text{ratio}=\frac{1/\sqrt2}{1-1/\sqrt2}=\frac{1}{\sqrt2-1}=\sqrt2+1 \).

Teacher's Note:
1. The particle does not move equal distances in equal times in SHM, so compute the positions at the end of each second.
2. T = 8 s means that 2 s is a quarter period.

Answer: (d) \( \frac{2\pi}{3}\sqrt{\frac mk} \)

Explanation:
1. The block performs SHM about the natural length with amplitude 2e until it reaches the wall at displacement e from the natural length.
2. From \( x=2e \): \( 2e\cos\omega t=e\Rightarrow\omega t=\frac\pi3 \), so the time to reach the wall is \( \frac{T_0}{6} \) with \( T_0=2\pi\sqrt{m/k} \).
3. After the elastic bounce it takes the same time to return: \( T=2\times\frac{T_0}{6}=\frac{T_0}{3}=\frac{2\pi}{3}\sqrt{\frac mk} \).

Teacher's Note:
1. The wall cuts off part of the SHM; the motion is periodic with a shorter period.
2. An elastic collision just reverses the velocity, so the second half is the mirror image of the first.

Answer: (c) \( T^2=t_1^2+t_2^2 \)

Explanation:
1. \( t\propto\frac{1}{\sqrt k}\Rightarrow k\propto\frac{1}{t^2} \).
2. Springs in series: \( \frac1k=\frac1{k_1}+\frac1{k_2} \), and \( k=\frac{4\pi^2m}{T^2} \) so \( T^2=t_1^2+t_2^2 \).

Teacher's Note:
1. Series combination: 1/k = 1/k1 + 1/k2. Since T^2 is proportional to 1/k, the squares of the periods add.
2. For parallel springs the reciprocals of the squares add instead.

Answer: (c) \( \sqrt{m/4K} \)

Explanation:
1. Left: \( 2K \) and \( 2K \) in series give \( K \).
2. Right: \( K \) and \( 2K \) in parallel give \( 3K \).
3. Springs on opposite sides of the mass act together: \( k_{eq}=K+3K=4K \).
4. \( T=2\pi\sqrt{\frac{m}{4K}} \).

Teacher's Note:
1. Reduce each side to one equivalent spring and then add the two sides.
2. Springs on opposite sides of a body add like springs in parallel.

Answer: (b) \( \frac65 \)

Explanation:
1. Acceleration of the suspension point: \( a=\frac{d^2y}{dt^2}=2k=2\ \text{m/s}^2 \) upward.
2. \( g_{eff}=g+a=12\ \text{m/s}^2 \).
3. \( \left(\frac{T_1}{T_2}\right)^2=\frac{g_{eff}}{g}=\frac{12}{10}=\frac65 \).

Teacher's Note:
1. Because T is proportional to 1/sqrt(g_eff), the square of the ratio of periods equals the ratio of effective gravities.
2. Upward acceleration of the support increases g_eff.

Answer: (a) \( \frac{3T}{4} \)

Explanation:
1. On one side of the swing the thread is free: effective length l, giving a half-oscillation of \( \frac T2 \).
2. On the other side the thread touches the rod and pivots about its lower end: the effective length is \( l-\frac{3l}{4}=\frac l4 \), whose period is \( 2\pi\sqrt{\frac{l/4}{g}}=\frac T2 \), giving a half-oscillation of \( \frac T4 \).
3. \( T'=\frac T2+\frac T4=\frac{3T}{4} \).

Teacher's Note:
1. Each half of the swing is half a period of a pendulum of the relevant length.
2. A quarter-length pendulum has half the period.

Answer: (d) \( \lt T \)

Explanation:
1. A uniform rod pivoted at one end has \( I=\frac{ml^2}{3} \) and centre of mass at \( \frac l2 \): \( T_{rod}=2\pi\sqrt{\frac{ml^2/3}{mg\,l/2}}=2\pi\sqrt{\frac{2l}{3g}} \).
2. The simple pendulum has \( T=2\pi\sqrt{l/g} \).
3. So \( T_{rod}=\sqrt{2/3}\,T\lt T \).

Teacher's Note:
1. Equivalent length of a uniform rod about its end is 2l/3.
2. Compare it with the simple pendulum of length l.

Answer: (d) \( \frac g{10} \)

Explanation:
1. Effective gravity \( g_{eff}=\sqrt{g^2+a^2}\approx g\left(1+\frac{a^2}{2g^2}\right) \).
2. \( \frac{T'}{T}=\left(\frac{g}{g_{eff}}\right)^{1/2}\approx1-\frac{a^2}{4g^2} \).
3. So \( \frac{3.99}{4}=0.9975=1-\frac{a^2}{4g^2}\Rightarrow\frac{a^2}{g^2}=0.01 \), giving \( a=0.1g=\frac g{10} \).

Teacher's Note:
1. In an accelerating car the effective gravity is the vector sum of g and the pseudo-force, with magnitude sqrt(g^2 + a^2).
2. Use the binomial approximation for a small change.

Answer: (c) 1.5 m

Explanation:
1. Radius \( R=1\ \text{m} \). Moment of inertia about a tangent in the plane of the ring: \( I=\frac12mR^2+mR^2=\frac32mR^2 \).
2. Equivalent length \( L=\frac{I}{md}=\frac{\frac32mR^2}{mR}=\frac32R=1.5\ \text{m} \).

Teacher's Note:
1. The moment of inertia of a ring about a diameter is half that about the perpendicular axis through its centre.
2. Add mR^2 using the parallel-axis theorem.

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

Explanation:
1. \( U=5x^2-20x\Rightarrow F=-\frac{dU}{dx}=-10x+20=-10(x-2) \).
2. The force is not constant ((a) false); it is a restoring force about x = 2 m, so the motion is SHM with maximum speed at x = 2 m ((b), (c) true).
3. \( k=10,\ T=2\pi\sqrt{\frac{0.1}{10}}=\frac{\pi}{5}\ \text{s} \) ((d) true).

Teacher's Note:
1. Force is minus the derivative of potential energy; a linear force with a negative slope means SHM.
2. The equilibrium is where U is minimum.

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

Explanation:
1. KE \( =E\left(1-\frac{x^2}{A^2}\right) \): for KE = 0.64E, \( x^2=0.36A^2 \), so x = 6 cm ((a) true).
2. At x = 5 cm, KE \( =E(1-0.25)=0.75E \), and the maximum PE equals E ((b) true).
3. Total energy is constant and equals the maximum KE ((c) true).
4. At x = A/2, \( v=\frac{\sqrt3}{2}v_{max} \), not half ((d) false).

Teacher's Note:
1. Total energy E = maximum KE = maximum PE.
2. KE = E (1 - x^2/A^2) and PE = E x^2/A^2.

Answer: (a) (a) and (b) only

Explanation:
1. \( x=A\sin\omega t=\frac A2\Rightarrow\omega t=\frac\pi6\Rightarrow t=\frac{T}{12} \) ((a) true, (d) false).
2. \( a=\omega^2x=\frac{\omega^2A}{2}=\frac{a_0}{2} \) ((b) true).
3. \( v=v_0\cos30^\circ=\frac{\sqrt3}{2}v_0 \), not \( v_0/2 \) ((c) false).

Teacher's Note:
1. Measure time from the mean position: x = A sin wt.
2. Acceleration is proportional to displacement.