JEE Physics Oscillations MCQs Set 18

Answer: (b) \( -\dfrac{4kx}{3} \)

Explanation:
1. Spring force: \( -2kx \). Let friction be f. Then \( f-2kx=Ma \).
2. Rolling without slipping (\( \alpha=a/R \)) with friction acting at the contact point gives \( f=-\tfrac12Ma \).
3. So \( -\tfrac12Ma-2kx=Ma \Rightarrow a=-\dfrac{4kx}{3M} \).
4. Net force \( =Ma=-\dfrac{4kx}{3} \).

Teacher's Note:
1. Net external force includes both the spring force and the friction force.
2. For rolling without slipping a quick method is energy: total energy is \( \tfrac12\cdot\tfrac32Mv^2+\tfrac12(2k)x^2 \).

Answer: (a) \( \sqrt{\dfrac{4k}{3M}} \)

Explanation:
1. Energy: \( E=\tfrac12Mv^2+\tfrac12I\omega_r^2+\tfrac12(2k)x^2=\tfrac34Mv^2+kx^2 \).
2. Differentiating, \( \tfrac32Mv\dot v+2kxv=0 \Rightarrow \ddot x=-\dfrac{4k}{3M}x \).
3. Hence \( \omega=\sqrt{4k/3M} \).

Teacher's Note:
1. A rolling disk has kinetic energy 3/4 Mv squared, because rotation adds half of the translational part.
2. The same effective mass 3M/2 works for any rolling solid disk.

Answer: (d) \( \mu g\sqrt{\dfrac{3M}{k}} \)

Explanation:
1. From \( f-2kx=Ma \) and \( a=-\tfrac{4kx}{3M} \): \( |f|=\tfrac{2kx}{3} \).
2. Friction is largest at the amplitude \( A=V_0/\omega \): \( \tfrac{2k}{3}A\le\mu Mg \).
3. \( V_0=\omega A\le\sqrt{\tfrac{4k}{3M}}\cdot\dfrac{3\mu Mg}{2k}=\mu g\sqrt{\dfrac{3M}{k}} \).

Teacher's Note:
1. Friction needed for rolling is greatest at the extreme positions where acceleration is largest.
2. Compare the required friction with its limiting value.

Answer: (c) \( -\dfrac{\sqrt3}{32}\pi^2 \) cm/s²

Explanation:
1. \( a=-\omega^2x \), with \( \omega=\pi/4 \) rad/s.
2. At \( t=4/3 \): \( x=\sin(\pi/3)=\tfrac{\sqrt3}{2} \) cm.
3. \( a=-\dfrac{\pi^2}{16}\cdot\dfrac{\sqrt3}{2}=-\dfrac{\sqrt3\pi^2}{32} \) cm/s².

Teacher's Note:
1. Acceleration in SHM is always directed opposite to the displacement.
2. Read the amplitude and period from the graph: amplitude 1 cm, period 8 s.

Answer: (b) \( \dfrac{1}{2\pi}\sqrt{\dfrac{6k}{M}} \)

Explanation:
1. Rotation by \( \theta \) stretches each spring by \( x=\theta L/2 \).
2. Restoring torque \( =2\cdot k\dfrac{L\theta}{2}\cdot\dfrac L2=\dfrac{kL^2}{2}\theta \).
3. \( I=\dfrac{ML^2}{12} \Rightarrow \omega^2=\dfrac{kL^2/2}{ML^2/12}=\dfrac{6k}{M} \).

Teacher's Note:
1. Both springs contribute torque in the same sense.
2. Use the moment of inertia of a rod about its centre, ML squared over 12.

Answer: (c) \( \dfrac1A\sqrt{\dfrac m\alpha} \)

Explanation:
1. Energy: \( \tfrac12mv^2+\alpha x^4=\alpha A^4 \), so \( v\sim A^2\sqrt{\alpha/m} \).
2. Time over a distance of order A: \( T\sim A/v\sim \dfrac1A\sqrt{\dfrac m\alpha} \).

Teacher's Note:
1. Dimensional reasoning: alpha has units of energy per length to the fourth.
2. The larger the amplitude, the faster the particle moves, so the period shrinks.

Answer: (a) (A) and (D)

Explanation:
1. The restoring torque is the torque of the weights about the pivot, the same in both cases for the same displacement, so (A) is true and (B) is false.
2. In case A the disk also rotates with the rod, so \( I_A=I_B+\tfrac12MR^2 \), larger than in case B.
3. \( \omega^2=\tau/(I\theta) \), so \( \omega_A\lt\omega_B \): (D) is true.

Teacher's Note:
1. Torque depends only on forces and the lever arms, not on the rotation of the disk.
2. More moment of inertia with the same torque gives a lower frequency.

Answer: (d) E1 = 4E2

Explanation:
1. For the same oscillator, \( E=\tfrac12m\omega^2A^2\propto A^2 \).
2. \( E_1/E_2=(2a/a)^2=4 \).

Teacher's Note:
1. The energy of an oscillator is proportional to the square of the amplitude.
2. The radius of a phase-space circle along the position axis is the amplitude.

Answer: (c) \( B=A,\ \phi=\dfrac{4\pi}{3} \)

Explanation:
1. Resultant of \( x_1,x_2 \): amplitude \( \sqrt{A^2+A^2+2A^2\cos120^\circ}=A \), phase \( \pi/3 \): \( x_{12}=A\sin(\omega t+\pi/3) \).
2. For rest, \( x_3=-x_{12} \): \( B=A \) and \( \phi=\pi/3+\pi=4\pi/3 \).

Teacher's Note:
1. Use phasor addition: two equal phasors 120 degrees apart give a phasor of the same length.
2. The third phasor must exactly cancel the resultant.

Answer: (b) \( \sqrt{50} \) m/s

Explanation:
1. Position of the block at t = 1 s: \( x=4.9+0.2\cos(\pi/3)=5.0 \) m from O.
2. Pebble covers 10 − 5 = 5 m horizontally in 1 s: \( v\cos45^\circ=5 \).
3. It must be at the block's height (zero) after 1 s: \( v\sin45^\circ\cdot1-\tfrac12(10)(1)^2=0\Rightarrow v\sin45^\circ=5 \).
4. \( v=5\sqrt2=\sqrt{50} \) m/s.

Teacher's Note:
1. Treat the horizontal and vertical motions of the pebble separately.
2. The block is at its mean position plus 0.2 cos(wt).

Answer: (a) (A) and (D)

Explanation:
1. Before the collision \( v=u_0\cos\omega t=0.5u_0 \Rightarrow \omega t=\pi/3 \). The collision reverses the velocity at the same position, so the energy and amplitude are unchanged: speed at equilibrium is \( u_0 \) (A true).
2. Then it needs another \( \pi/3 \) to reach equilibrium: first passage at \( \omega t=2\pi/3 \), i.e. \( t=\tfrac{2\pi}{3}\sqrt{m/k} \). So (B) is false.
3. It then goes to the far extreme (extra \( \pi/2 \)) at \( \omega t=7\pi/6 \) and returns to equilibrium at \( \omega t=7\pi/6+\pi/2=5\pi/3 \): (D) true, (C) false.

Teacher's Note:
1. The wall only changes the direction of the velocity here, so the energy is conserved.
2. Track the phase angle on the reference circle.

Answer: (d) (B) and (D)

Explanation:
1. Momentum amplitude \( =m\omega\times \) position amplitude. Ellipse: \( b=m\omega_1a \). Circle: \( R=m\omega_2R\Rightarrow m\omega_2=1 \).
2. \( \omega_2/\omega_1=a\,m\omega_2/b=a/b=n^2 \): (B) true. (C) would need \( \omega_1\omega_2=n^2 \); here \( \omega_1\omega_2=1/(n^2m^2) \): false.
3. \( E_1=b^2/2m,\ E_2=R^2/2m \). \( E_1/\omega_1=\dfrac{ab}{2} \) and \( E_2/\omega_2=\dfrac{R^2}{2}=\dfrac{a^2}{2n^2}=\dfrac{ab}{2} \) since \( b=a/n^2 \): (D) true, (A) false.

Teacher's Note:
1. Energy over frequency is the action of the oscillator, an adiabatic invariant.
2. Check each relation using the semi-axes in terms of a and n.

Answer: (c) (A), (B) and (D)

Explanation:
1. (i) At \( x_0 \) momentum conservation: \( v'=Mv/(M+m) \). New \( \omega'=\sqrt{k/(M+m)} \), so the amplitude \( v'/\omega'=A\sqrt{M/(M+m)} \). (ii) At the extreme the speed is zero, so the amplitude is unchanged: (A) true.
2. Time period \( 2\pi\sqrt{(M+m)/k} \) in both: (B) true.
3. Energy decreases in (i) but is unchanged in (ii): (C) false.
4. Speed at \( x_0 \) is \( A'\omega' \): in (i) it is \( v'\lt v \); in (ii) it is \( A\omega'\lt A\omega \): (D) true.

Teacher's Note:
1. A soft collision at the extreme position costs no energy because nothing is moving.
2. The new frequency depends only on the total mass and k.

Answer: (a) 0.01 s

Explanation:
1. \( v_{max}=A\omega=2\pi A/T \).
2. \( T=2\pi(7\times10^{-3})/4.4=0.01 \) s.

Teacher's Note:
1. Maximum speed equals amplitude times angular frequency.
2. Convert millimetres to metres first.

Answer: (d) 0.5 s

Explanation:
1. Speed is maximum at the mean position, \( x=0 \).
2. \( \cos\pi t=0\Rightarrow \pi t=\pi/2\Rightarrow t=0.5 \) s.

Teacher's Note:
1. Maximum speed occurs when displacement is zero.
2. The first zero of the cosine is at a quarter cycle.

Answer: (c) \( A=x_0\omega^2,\ \delta=3\pi/4 \)

Explanation:
1. \( a=-\omega^2x_0\cos(\omega t-\pi/4)=\omega^2x_0\cos(\omega t-\pi/4+\pi) \).
2. So \( A=x_0\omega^2 \) and \( \delta=3\pi/4 \).

Teacher's Note:
1. The minus sign is the same as adding pi to the phase.
2. Acceleration has amplitude omega squared times the displacement amplitude.

Answer: (b) 2f

Explanation:
1. The two springs act in parallel: \( k=k_1+k_2 \), \( f=\dfrac1{2\pi}\sqrt{k/m} \).
2. Both quadrupled: \( k\to4k \), so \( f\to2f \).

Teacher's Note:
1. A mass between two springs sees the sum of the stiffnesses.
2. Frequency goes as the square root of k.

Answer: (d) (A) and (D)

Explanation:
1. \( a=-\omega^2x \), \( \omega=2\pi/T \): \( aT/x=-\omega^2T=-2\pi\omega \), constant (A).
2. \( v^2=\omega^2(A^2-x^2) \): \( a^2T^2+4\pi^2v^2=4\pi^2\omega^2x^2+4\pi^2\omega^2(A^2-x^2)=4\pi^2\omega^2A^2 \), constant (D).
3. \( aT+2\pi v \) and \( aT/v \) vary with time.

Teacher's Note:
1. Substitute the SHM relations and see whether x or v remains.
2. Combinations that reduce to the amplitude and omega are constants.

Answer: (c) \( \left(\dfrac{M+m}{M}\right)^{1/2} \)

Explanation:
1. Momentum conservation: \( Mv=(M+m)v' \), \( v=A_1\omega \), \( v'=A_2\omega' \).
2. \( \omega'=\omega\sqrt{M/(M+m)} \).
3. \( A_2=\dfrac{M}{M+m}\cdot\dfrac{\omega}{\omega'}A_1=A_1\sqrt{\dfrac M{M+m}} \), hence \( A_1/A_2=\sqrt{(M+m)/M} \).

Teacher's Note:
1. Energy is lost in the sticking collision, momentum is not.
2. The new frequency is lower because the mass is greater.

Answer: (b) \( \dfrac{5k}{2} \)

Explanation:
1. Stiffness is inversely proportional to the length: \( k\ell=\text{constant} \).
2. Part A is \( 2/5 \) of the length, so \( k_A=\dfrac{k}{2/5}=\dfrac{5k}{2} \).

Teacher's Note:
1. A shorter piece of spring is stiffer.
2. Fractional length of A is 2 out of 5.

Answer: (d) \( 2/b \)

Explanation:
1. Equation of motion \( \ddot x=-\omega^2x-b\dot x \), in the form \( \ddot x+b\dot x+\omega^2x=0 \): amplitude \( A=A_0e^{-bt/2} \).
2. \( A_0/e=A_0e^{-b\tau/2}\Rightarrow\tau=2/b \).

Teacher's Note:
1. Damping with force proportional to velocity gives an exponentially falling amplitude.
2. Here b is the damping per unit mass.