JEE Physics Oscillations MCQs Set 19

Answer: (c) 0.729

Explanation:
1. \( A=A_0e^{-\gamma t} \): \( e^{-5\gamma}=0.9 \).
2. At \( t=15 \) s: \( \alpha=e^{-15\gamma}=0.9^3=0.729 \).

Teacher's Note:
1. Equal time intervals multiply the amplitude by equal factors.
2. 15 s is three intervals of 5 s.

Answer: (a) \( \dfrac{7A}{3} \)

Explanation:
1. Before: \( v^2=\omega^2(A^2-4A^2/9)=\tfrac59\omega^2A^2 \).
2. After tripling: \( v'^2=9v^2=5\omega^2A^2 \).
3. \( A'^2=x^2+v'^2/\omega^2=\tfrac49A^2+5A^2=\tfrac{49}{9}A^2\Rightarrow A'=\dfrac{7A}{3} \).

Teacher's Note:
1. Frequency is unchanged; only the amplitude changes.
2. Use v squared equals omega squared times (A squared minus x squared).

Answer: (d) 7.1 N/m

Explanation:
1. \( m=\dfrac{108\times10^{-3}}{6.02\times10^{23}}=1.79\times10^{-25} \) kg.
2. \( k=m(2\pi f)^2=1.79\times10^{-25}\times(6.28\times10^{12})^2\approx7.1 \) N/m.

Teacher's Note:
1. Find the mass of one atom first.
2. Use k equals m omega squared.

Answer: (c) 0.20

Explanation:
1. Period 0.20 s gives \( \omega=10\pi \) rad/s, \( \omega^2=100\pi^2=1000 \).
2. \( m=k/\omega^2=200/1000=0.20 \) kg.

Teacher's Note:
1. Read the period from the velocity function.
2. Mass follows from k over omega squared.

Answer: (b) −0.20 m

Explanation:
1. Amplitude \( A=v_{max}/\omega=2\pi/10\pi=0.20 \) m.
2. At t = 0 the speed is zero (an extreme position) and it is just starting to move in the +x direction, so it is at \( x=-0.20 \) m.

Teacher's Note:
1. Velocity is zero at extreme positions.
2. A positive velocity just after t = 0 means the block started from the negative extreme.

Answer: (a) −200

Explanation:
1. \( a=dv/dt=20\pi^2\cos(10\pi t)=200\cos(10\pi t) \).
2. At t = 0.10 s: \( \cos\pi=-1\Rightarrow a=-200 \) m/s².

Teacher's Note:
1. Differentiate the velocity function.
2. At half a period the block is at the opposite extreme.

Answer: (d) 4.0

Explanation:
1. \( KE_{max}=\tfrac12mv_{max}^2=\tfrac12(0.20)(2\pi)^2=0.4\pi^2=4.0 \) J.

Teacher's Note:
1. Kinetic energy is greatest at the mean position.
2. Use pi squared equals 10 as instructed.

Answer: (c) 6.93 s

Explanation:
1. Amplitude \( A=A_0e^{-bt/2m} \).
2. \( \tfrac12=e^{-bt/2m}\Rightarrow t=\dfrac{2m\ln2}{b}=\dfrac{2(0.2)(0.693)}{0.04}=6.93 \) s.

Teacher's Note:
1. Use SI units: 40 g/s is 0.04 kg/s.
2. The amplitude decays exponentially at rate b over 2m.

Answer: (b) 3.47 s

Explanation:
1. Energy \( E=E_0e^{-bt/m} \).
2. \( \tfrac12=e^{-bt/m}\Rightarrow t=\dfrac{m\ln2}{b}=\dfrac{0.2\times0.693}{0.04}=3.47 \) s, half of the amplitude half-life.

Teacher's Note:
1. Energy goes as the square of the amplitude, so it decays twice as fast.
2. Use the mass in kilograms.

Answer: (a) 0.30 s

Explanation:
1. \( \omega'=\sqrt{k/m-b^2/4m^2}=\sqrt{450-0.01}\approx21.2 \) rad/s.
2. \( T=2\pi/\omega'=0.296\approx0.30 \) s.

Teacher's Note:
1. Light damping barely changes the frequency.
2. The damping term is tiny compared with k over m.

Answer: (d) \( \dfrac{\sqrt{11}}{2} \)

Explanation:
1. Compare with \( m\ddot x+b\dot x+kx=0 \): m = 1, b = 1, k = 3. \( b^2\lt 4mk \) so it is underdamped.
2. \( \omega_d=\sqrt{k/m-b^2/4m^2}=\sqrt{3-\tfrac14}=\dfrac{\sqrt{11}}{2} \).

Teacher's Note:
1. Underdamping needs b squared less than 4mk.
2. Subtract the damping term inside the root.

Answer: (b) \( \dfrac{1}{m(\omega_0^2-\omega^2)} \)

Explanation:
1. For small damping the forced amplitude is \( A=\dfrac{F_0}{m|\omega_0^2-\omega^2|} \).
2. So \( A\propto\dfrac{1}{m(\omega_0^2-\omega^2)} \).

Teacher's Note:
1. The amplitude becomes large near resonance.
2. Amplitude is force divided by m times the difference of squared frequencies.

Answer: (a) V/2

Explanation:
1. Maximum speed is where \( F=Kx \), \( x=F/K \). Work-energy: \( Fx-\tfrac12Kx^2=\tfrac12mV^2\Rightarrow V^2=F^2/(mK) \).
2. With 4K: \( V'^2=F^2/(4mK)=V^2/4 \), so \( V'=V/2 \).

Teacher's Note:
1. Speed is greatest where the net force is zero.
2. Replace K by 4K in the formula.

Answer: (d) \( \pi \)

Explanation:
1. \( I=m(1^2+3^2)=10m \); restoring torque \( \approx mg(1+3)\theta=4mg\theta \).
2. \( T=2\pi\sqrt{\dfrac{10m}{4mg}}=2\pi\sqrt{\dfrac{10}{40}}=\pi \) s.

Teacher's Note:
1. The rod is light, so only the masses contribute to the moment of inertia.
2. Use the physical pendulum formula with the torque of both weights.

Answer: (b) 25

Explanation:
1. Restoring torque for small \( \theta \): \( mgL\theta+k\ell\theta\cdot\ell=mgL\theta+9mgL\theta=10mgL\theta \). \( I=mL^2 \).
2. \( \omega^2=10g/L=100/L \), \( T=2\pi\sqrt{L/100}=\pi\sqrt{L/25} \).
3. So n = 25.

Teacher's Note:
1. Both the weight and the spring give restoring torque.
2. Write the period as pi times a root to compare.

Answer: (a) \( \dfrac8\pi \) m/s

Explanation:
1. The period is 4 s; the velocity keeps one sign for 2 s, and the interval contains \( 2n-1 \) such half periods.
2. Distance in one half period: \( \int_0^2 4\sin(\pi t/2)\,dt=\dfrac{16}{\pi} \) m.
3. Average speed \( =\dfrac{(2n-1)\,16/\pi}{2(2n-1)}=\dfrac8\pi \) m/s.

Teacher's Note:
1. Speed is the magnitude of velocity, so every half period adds distance.
2. The answer does not depend on n.

Answer: (d) \( \dfrac\pi3\sqrt{\dfrac\ell g} \)

Explanation:
1. With \( \omega=\sqrt{g/\ell} \): A has angle \( -\theta\cos\omega t \) and B (length \( \ell/4 \), frequency \( 2\omega \)) has \( \theta\cos2\omega t \).
2. Parallel strings: equal angles, \( \cos2\omega t+\cos\omega t=0\Rightarrow2\cos\tfrac{3\omega t}{2}\cos\tfrac{\omega t}2=0 \).
3. First solution: \( \tfrac{3\omega t}{2}=\tfrac\pi2\Rightarrow t=\dfrac{\pi}{3\omega}=\dfrac\pi3\sqrt{\dfrac\ell g} \).

Teacher's Note:
1. Shorter pendulums swing faster: B has twice the frequency of A.
2. Strings are parallel when their angles from the vertical are equal.

Answer: (b) 2.83

Explanation:
1. Time to cross the gap each way: \( 0.60/1.20=0.50 \) s; two crossings per cycle: 1.00 s.
2. Half period on spring 1: \( \pi\sqrt{m/k_1}=\pi\sqrt{0.2/1.8}=\pi/3=1.047 \) s; on spring 2: \( \pi\sqrt{0.2/3.2}=\pi/4=0.785 \) s.
3. \( T=1.00+1.047+0.785\approx2.83 \) s.

Teacher's Note:
1. The block moves freely in the gap and as SHM while touching a spring.
2. Add the free-motion time and the two half periods.

Answer: (a) \( 2\pi\sqrt{\dfrac{\ell}{\mu g}} \)

Explanation:
1. If the plank is displaced by x, the normal reactions are \( N_1=\dfrac{Mg(\ell-x)}{2\ell} \) and \( N_2=\dfrac{Mg(\ell+x)}{2\ell} \).
2. Net friction \( =\mu(N_2-N_1)=\dfrac{\mu Mg}{\ell}x \), directed towards the centre.
3. \( \omega^2=\dfrac{\mu g}{\ell} \Rightarrow T=2\pi\sqrt{\dfrac{\ell}{\mu g}} \).

Teacher's Note:
1. The wheels turn so that friction always points towards the centre of the plank.
2. Without the coefficient of friction the answer would be incomplete.

Answer: (d) 0.40 J

Explanation:
1. Displacement of the centre of mass \( =\dfrac{1(0.1)+2(0.05)}{3}=\dfrac{0.2}{3} \) m.
2. Only the external force changes the centre-of-mass energy: \( KE_{cm}=F\Delta x_{cm}=6\times\dfrac{0.2}{3}=0.40 \) J.

Teacher's Note:
1. The internal spring forces do not change the motion of the centre of mass.
2. Use the work-energy theorem for the centre of mass.

Answer: (c) 0.26 s

Explanation:
1. \( KE_{cm}=F\Delta x_{cm}=6\times\tfrac{0.2}{3}=0.40 \) J, so \( v_{cm}=\sqrt{2(0.40)/3}=0.516 \) m/s.
2. Impulse: \( Ft=(m_1+m_2)v_{cm}\Rightarrow t=\dfrac{3(0.516)}{6}=0.26 \) s.

Teacher's Note:
1. Impulse equals the change in momentum of the whole system.
2. The centre of mass accelerates at F over the total mass.