JEE Physics Oscillations MCQs Set 09

Answer: (a) displacement from the mean position

Explanation:
1. In SHM the restoring force is \( F=-kx \), so \( a=-\omega^2x \).
2. Acceleration is therefore proportional to the (signed) displacement from the mean position and directed towards it.

Teacher's Note:
1. Displacement is a signed quantity measured from the mean position, while distance is always positive.
2. The negative sign in a = -w^2 x is what makes the force restoring.

Answer: (d) the ratio of their maximum speeds is 1 : 1

Explanation:
1. First: \( y=A+A\cos2\omega t \) is SHM about y = A with amplitude A and angular frequency 2\( \omega \). \( v_{max}=2A\omega,\ a_{max}=4A\omega^2 \).
2. Second: \( y=2A\sin(\omega t+60^\circ) \) is SHM with amplitude 2A and angular frequency \( \omega \). \( v_{max}=2A\omega,\ a_{max}=2A\omega^2 \).
3. Both are SHM, the maximum speeds are equal (1 : 1) and the accelerations are in the ratio 2 : 1.

Teacher's Note:
1. cos^2(wt) can be rewritten with a double-angle identity to show it is a shifted SHM of double frequency.
2. a sin(wt) + b cos(wt) always combines into one sine of amplitude sqrt(a^2+b^2).

Answer: (c) \( 0.15\sqrt3\ \text{m/s} \)

Explanation:
1. Amplitude \( A=0.15\ \text{m} \), \( \omega=2\ \text{rad/s} \), \( y=0.075\ \text{m} \).
2. \( v=\omega\sqrt{A^2-y^2}=2\sqrt{0.0225-0.005625}=2(0.075\sqrt3) \) \( =0.15\sqrt3\ \text{m/s} \).

Teacher's Note:
1. Use v = w sqrt(A^2 - x^2) when speed at a given displacement is asked.
2. Keep all lengths in metres before substituting.

Answer: (b) \( \frac32 \)

Explanation:
1. Rewrite as \( \frac{d^2y}{dt^2}+\frac94y=0 \).
2. Comparing with \( \frac{d^2y}{dt^2}+\omega^2y=0 \) gives \( \omega^2=\frac94 \), so \( \omega=\frac32\ \text{rad/s} \).

Teacher's Note:
1. The standard SHM equation is y'' + w^2 y = 0, so divide through to make the coefficient of y'' equal to one.
2. Note that w^2 equals 9/4, not w itself.

Answer: (b) an extreme position

Explanation:
1. At the mean position the particle passes every T/2 (in alternate directions).
2. At any other interior point the gaps between successive passages alternate between two unequal values that add up to T.
3. Only at an extreme position does the particle appear once per cycle, so the gap is exactly T.

Teacher's Note:
1. An interior point is visited twice in every cycle, once in each direction.
2. The extreme is visited just once per cycle.

Answer: (a) A = 5, T = 2

Explanation:
1. Comparing with \( y=A\sin(\omega t+\phi) \): \( A=5\ \text{m},\ \omega=\pi\ \text{rad/s} \).
2. \( T=2\pi/\omega=2\ \text{s} \).

Teacher's Note:
1. Read amplitude as the coefficient of the sine and w as the coefficient of t.
2. The constant 4pi inside the bracket is only the initial phase.

Answer: (d) \( \frac{3}{2\pi}\ \text{m} \)

Explanation:
1. \( \omega=\frac{2\pi}{6}=\frac{\pi}{3}\ \text{rad/s} \).
2. At t = 2 s the particle is 1 s past the mean position, so its phase from the mean position is \( \omega\times1=\pi/3 \).
3. \( v=A\omega\cos\frac{\pi}{3}=\frac{A\omega}{2}=0.25\Rightarrow A\omega=0.5 \).
4. \( A=\frac{0.5}{\pi/3}=\frac{3}{2\pi}\ \text{m} \).

Teacher's Note:
1. Measure phase from the mean position: x = A sin(wt), v = A w cos(wt).
2. Here t is counted from the instant the particle is at the mean position.

Answer: (c) \( 3a/T \)

Explanation:
1. Take \( x=a\cos\omega t \) from the extreme. The particle reaches \( x=a/2 \) when \( \cos\omega t=\tfrac12 \), i.e. \( \omega t=\pi/3 \), so \( t=T/6 \).
2. \( v_{mean}=\frac{a/2}{T/6}=\frac{3a}{T} \).

Teacher's Note:
1. Mean velocity = displacement / time, with the displacement here equal to a/2.
2. Reference-circle angle of 60 degrees corresponds to one sixth of the period.

Answer: (c) \( 135^\circ \)

Explanation:
1. First particle: \( x_1=A\sin\phi_1=A/\sqrt2 \) and moving towards the mean, so \( \phi_1=135^\circ \).
2. Second particle: \( x_2=2A\sin\phi_2=-2A \) so \( \phi_2=270^\circ \).
3. Phase difference \( 270^\circ-135^\circ=135^\circ \).

Teacher's Note:
1. Find each phase from both position and direction of motion (sign of velocity).
2. The extreme on the opposite side corresponds to phase 270 degrees in x = A sin(phi).

Answer: (b) \( T/6 \)

Explanation:
1. Particle 1 starts at the extreme: \( y_1=A\cos\omega t \).
2. Particle 2 starts at \( -A/2 \) moving towards the positive side: \( y_2=A\cos(\omega t+240^\circ) \).
3. They cross when \( \cos\omega t=\cos(\omega t+240^\circ)\Rightarrow \omega t+240^\circ=360^\circ-\omega t \), so \( \omega t=60^\circ \) and \( t=T/6 \).

Teacher's Note:
1. Write both displacements as cosines with the correct initial phases using direction of motion.
2. Check: at t = T/6 both are at +A/2.

Answer: (a) \( \sqrt{k/m_2} \)

Explanation:
1. After removing \( m_1 \) only \( m_2 \) hangs on the spring.
2. The angular frequency depends only on the spring constant and the mass that is oscillating: \( \omega=\sqrt{k/m_2} \).

Teacher's Note:
1. Removing m1 changes the equilibrium position (and sets the amplitude), but not the angular frequency of m2.
2. The period of a vertical spring-mass system does not depend on g.

Answer: (d) 0.133

Explanation:
1. \( \omega^2=\frac{K}{M+m}=\frac{20}{6}\ \text{s}^{-2} \).
2. \( a_{max}=\omega^2A=\frac{20}{6}(0.4)=\frac43\ \text{m/s}^2 \).
3. The friction on the upper block must supply this: \( \mu mg\ge m\,a_{max}\Rightarrow\mu\ge\frac{4/3}{10}=0.133 \).

Teacher's Note:
1. Both blocks oscillate together, so use the total mass in omega squared.
2. The upper block is on the verge of slipping at the extreme where acceleration is greatest.

Answer: (d) \( \frac{1}{\frac{2L}{v}+\pi\sqrt{\frac mK}} \)

Explanation:
1. Equal masses colliding elastically exchange velocities. The first block stops and the spring block moves off with speed v.
2. The spring block performs half an oscillation, \( \frac T2=\pi\sqrt{m/K} \), and returns with speed v; it collides again and hands the velocity back, so the first block now moves left with speed v.
3. The first block goes to the left wall and back, taking \( \frac{2L}{v} \).
4. One cycle between successive collisions pairs: \( t=\frac{2L}{v}+\pi\sqrt{\frac mK} \), so the frequency is \( \frac1t \).

Teacher's Note:
1. Elastic collision of equal masses = exchange of velocities.
2. The spring block spends half a period in contact-free SHM, and the free block spends 2L/v travelling to the wall and back.

Answer: (c) \( \frac{kA}{2} \)

Explanation:
1. Together the blocks have mass \( 2m \), so \( a_{max}=\frac{kA}{2m} \).
2. The only horizontal force on P is friction: \( f_{max}=m\,a_{max}=m\cdot\frac{kA}{2m}=\frac{kA}{2} \).

Teacher's Note:
1. Find the common acceleration first, then apply Newton's second law to just the block whose friction is asked.
2. Friction is static and need not equal mu_s m g.

Answer: (b) \( 2\pi\sqrt{\frac{3m}{2k}} \)

Explanation:
1. The lower pair in parallel: \( k+k=2k \).
2. In series with the upper spring: \( k_{eq}=\frac{k\cdot2k}{k+2k}=\frac{2k}{3} \).
3. \( T=2\pi\sqrt{\frac{m}{k_{eq}}}=2\pi\sqrt{\frac{3m}{2k}} \).

Teacher's Note:
1. First reduce each parallel group, then combine the groups in series.
2. The same force passes through springs in series; the same extension acts on springs in parallel.

Answer: (a) \( \frac1{2\pi}\sqrt{\frac{K_1+K_2}{m}} \)

Explanation:
1. When the block moves by x, one spring is stretched and the other compressed by x, and both push the block the same way.
2. Restoring force \( F=-(K_1+K_2)x \), so \( k_{eq}=K_1+K_2 \).
3. \( f=\frac1{2\pi}\sqrt{\frac{K_1+K_2}{m}} \).

Teacher's Note:
1. Springs on opposite sides of a block behave like springs in parallel.
2. Springs in a straight line end to end behave like springs in series.

Answer: (a) 4k

Explanation:
1. Whichever way the mass is displaced, all three springs produce restoring forces in the same direction (one is stretched while the others are compressed, or vice versa).
2. So they act as springs in parallel: \( k_{eq}=2k+k+k=4k \).

Teacher's Note:
1. Springs attached to the same body on both of its sides all oppose its displacement together, so they add like parallel springs.
2. Do not use the series rule unless the springs share the same force.

Answer: (d) \( T_s \) remains the same, \( T_p \) decreases

Explanation:
1. Spring-block period \( 2\pi\sqrt{m/k} \) does not depend on g, so \( T_s \) stays the same.
2. For an upward-accelerating lift, \( g_{eff}=g+a \) is larger, so \( T_p=2\pi\sqrt{l/g_{eff}} \) decreases.

Teacher's Note:
1. Only the pendulum depends on the effective gravity.
2. In a lift accelerating upwards a body feels heavier, so a pendulum swings faster.

Answer: (c) \( T_1\gt T_2 \)

Explanation:
1. Constant velocity: no acceleration, so \( g_{eff}=g \) and \( T_1=2\pi\sqrt{l/g} \).
2. Moving down while retarding means the acceleration is upwards: \( g_{eff}=g+a\gt g \), so \( T_2=2\pi\sqrt{l/(g+a)}\lt T_1 \).

Teacher's Note:
1. A lift slowing down while descending has an upward acceleration.
2. Period does not depend on the mass of the bob.

Answer: (b) \( mg\left[1+\left(\frac Al\right)^2\right] \)

Explanation:
1. Tension is maximum at the lowest point, where \( T-mg=\frac{mv^2}{l} \).
2. Maximum speed \( v=\omega A=\sqrt{\frac gl}A \) so \( v^2=\frac{gA^2}{l} \).
3. \( T_{max}=mg+\frac{m}{l}\cdot\frac{gA^2}{l}=mg\left[1+\frac{A^2}{l^2}\right] \).

Teacher's Note:
1. At the bottom the string supports the weight and supplies the centripetal force.
2. A here is the linear amplitude, so A/l is the angular amplitude in radians.

Answer: (b) 5

Explanation:
1. They are in phase again when the numbers of oscillations n (period T) and m (period 5T/4) completed in the same time satisfy \( nT=m\cdot\frac{5T}{4} \).
2. \( n=\frac{5m}{4} \). The smallest integer solution is \( m=4,\ n=5 \).
3. So the pendulum with period T has made 5 oscillations.

Teacher's Note:
1. The pendulums are in phase again when both have completed a whole number of oscillations at the same instant.
2. Find the least common multiple of the two periods.

Answer: (a) first increases and then decreases

Explanation:
1. The effective length is the distance from the point of suspension to the centre of mass of the sphere with its water.
2. As water drains, the centre of mass first moves lower, so the effective length increases and the period increases.
3. When most of the water is gone the centre of mass rises back to the centre of the shell, so the effective length and the period decrease again.

Teacher's Note:
1. The period of a pendulum depends on the position of the centre of mass of the bob, not on its mass.
2. The centre of mass dips and then returns to the geometric centre of the shell.

Answer: (d) \( \frac{a^2-x^2}{x^2} \)

Explanation:
1. \( T=\tfrac12m\omega^2(a^2-x^2) \) and \( V=\tfrac12m\omega^2x^2 \).
2. \( \frac TV=\frac{a^2-x^2}{x^2} \).

Teacher's Note:
1. Kinetic energy is (1/2) m w^2 (a^2 - x^2) and potential energy is (1/2) m w^2 x^2.
2. The two always add up to the constant total energy (1/2) m w^2 a^2.

Answer: (c) 0.5 m

Explanation:
1. About a point on the rim, in its own plane: \( I=2mR^2 \) and the distance of the centre of mass from the pivot is R, so \( T=2\pi\sqrt{\frac{2mR^2}{mgR}}=2\pi\sqrt{\frac{2R}{g}} \).
2. A seconds pendulum has \( T=2\ \text{s} \): \( \frac{2R}{g}=\frac1{\pi^2}\Rightarrow R=\frac{g}{2\pi^2}=0.5\ \text{m} \).

Teacher's Note:
1. A ring pivoted at its rim in its own plane acts like a simple pendulum of length 2R.
2. A seconds pendulum has a period of exactly 2 s.

Answer: (c) \( \sqrt{\frac{12C}{ml^2}} \)

Explanation:
1. Torsional oscillation: \( \omega=\sqrt{C/I} \).
2. For a rod about a vertical axis through its centre \( I=\frac{ml^2}{12} \).
3. \( \omega=\sqrt{\frac{12C}{ml^2}} \), so the frequency is proportional to this.

Teacher's Note:
1. For a torsion pendulum the restoring torque is -C theta and the equation is I theta'' = -C theta.
2. Moment of inertia of a rod about its centre perpendicular to its length is ml^2/12.