JEE Physics Oscillations MCQs Set 25

Answer: (d) 5T

Explanation:
1. If the time is \( t \), then \( t \) = nT = \( m(5T/4) \) with the smallest integers: \( n = 5 \) and \( m = 4 \).
2. So \( t = 5T \).

Teacher's Note:
1. Least common multiple of the periods.

Answer: (b) \( 2\sqrt2 \)

Explanation:
1. \( g ∝ M/R^{2} \), so g′ = \( g \times 2/4 = g/2 \).
2. T′ = 2π√(l/g′) = \( \sqrt{2} \times 2 s = 2\sqrt{2} s \).

Teacher's Note:
1. g′ = \( g/2 \).

Answer: (d) 10%

Explanation:
1. \( T ∝ \sqrt{l} \), so T′/T = \( \sqrt{1.21} = 1.1 \).
2. The time period increases by 10%.

Teacher's Note:
1. \( T ∝ \sqrt{l} \).

Answer: (c) the amplitude of oscillation should be small

Explanation:
1. The motion is simple harmonic only if \( \sin \theta \approx \theta \).
2. That needs a small angular amplitude.

Teacher's Note:
1. Small angle approximation.

Answer: (d) time period of oscillations is 6τ

Explanation:
1. \( x = A(1 - \cos \omegat) \): A(1 − cos ωτ) = \( a \) and A(1 − cos 2ωτ) = \( 3a \).
2. Dividing gives \( 1 + \cos \) ωτ = 3/2, so ωτ = π/3 and \( T \) = 6τ (the amplitude is \( 2a \)).

Teacher's Note:
1. Use the two distances to find ωτ.

Answer: (b) 50 J

Explanation:
1. PE = \( E(x/A)^{2} = 100 \times 0.5 = 50 J \).
2. KE = \( 100 - 50 = 50 J \).

Teacher's Note:
1. KE = \( E(1 - x^{2}/A^{2}) \).

Answer: (b) The effective spring constant K of springs in parallel is given by 1/K = 1/K₁ + 1/K₂ + …

Explanation:
1. For springs in parallel the spring constants add, \( K = K_{1} + K_{2} \) + ….
2. The formula in \( (b) \) is for springs in series, so \( (b) \) is wrong.

Teacher's Note:
1. Parallel: add \( k \).
2. Series: add \( 1/k \).

Answer: (c) III only

Explanation:
1. In one full vibration the particle passes the mean position twice (KE maximum) and each extreme once (PE maximum, twice in all).
2. So both become maximum twice.

Teacher's Note:
1. Two extremes, two crossings.

Answer: (a) \( \pi\sqrt2\ \text{s} \)

Explanation:
1. Comparing with \( d^{2}y/dt^{2} + \omega^{2}y = 0 \) gives \( \omega^{2} = 2 \).
2. \( T = 2\pi/\sqrt{2} = \pi\sqrt{2} s \).

Teacher's Note:
1. ω² is the coefficient of \( y \).

Answer: (a) \( \langle E\rangle=\langle U\rangle \)

Explanation:
1. Averaged over a period, ⟨sin²⟩ = ⟨cos²⟩ = 1/2.
2. So both energies average to half the total energy.

Teacher's Note:
1. Equal averages.

Answer: (c) A-3, B-4, C-2, D-3

Explanation:
1. \( F = 0 \) at \( x = 4 (A-3) \).
2. Released from rest at \( x = 6 \), so \( A = 2 (B-4) \).
3. \( \omega = \sqrt{2/2} = 1 \) rad/s, and going from \( x = 2 \) to \( x = 4 \) takes \( T/4 = \pi/2 (C-2) \).
4. \( E \) = ½kA² = \( \frac{1}{2} \times 2 \times 4 = 4 J (D-3) \).

Teacher's Note:
1. Locate the mean position first.

Answer: (b) decrease

Explanation:
1. The bob is attracted downwards by the plate, adding to gravity.
2. The effective \( g \) increases, so \( T = 2\pi\sqrt{l/g} \) decreases.

Teacher's Note:
1. Extra downward force.

Answer: (c) 0.729

Explanation:
1. A = \( A_{0}e^{-\lambdat} \), so the amplitude falls by the factor 0.9 every \( 5 s \).
2. In another \( 10 s \) (two such intervals) it falls to 0.9² of the value at \( 5 s \): the total factor is \( 0.9^{3} = 0.729 \) of the original.

Teacher's Note:
1. Exponential decay.

Answer: (d) square of the amplitude of motion

Explanation:
1. \( E = \frac{1}{2}m\omega^{2}A^{2} \).
2. So it is proportional to the square of the amplitude (and, at the same time, also to the square of the velocity there).

Teacher's Note:
1. \( E ∝ A^{2} \).

Answer: (a) \( U_x=\frac k2(x-a)^2 \)

Explanation:
1. SHM needs a potential energy that is quadratic about the equilibrium position.
2. Only \( (k/2)(x - a)^{2} \) has this form.

Teacher's Note:
1. \( U ∝ (x - x_{0})^{2} \).

Answer: (c) \( \frac{2\pi\beta}{\alpha} \)

Explanation:
1. \( \alpha = \omega^{2}A \) and \( \beta = \omegaA \), so \( \omega = \alpha/\beta \).
2. \( T = 2\pi/\omega = 2\pi\beta/\alpha \).

Teacher's Note:
1. Divide the two maxima.

Answer: (a) \( t=2t_0 \)

Explanation:
1. In water the effective g′ = g(1 − ρ_w/ρ_b) = \( g(1 - 3/4) = g/4 \).
2. \( T ∝ \) 1/√g′, so \( t = 2t_{0} \).

Teacher's Note:
1. Buoyancy reduces the effective \( g \).

Answer: (b) \( \pi^2ma^2\nu^2 \)

Explanation:
1. The average KE over a quarter period is \( \frac{1}{4}m\omega^{2}a^{2} \).
2. With ω = 2πν this equals π²ma²ν².

Teacher's Note:
1. ⟨KE⟩ = \( E/2 = \frac{1}{4}m\omega^{2}a^{2} \).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( T = 2\pi\sqrt{m/k} \), so a larger \( k \) gives a smaller \( T \).
2. The reason is the explanation.

Teacher's Note:
1. \( T ∝ 1/\sqrt{k} \).

Answer: (c) A is true, but R is false.

Explanation:
1. At the extremes the velocity is zero, so the assertion is true.
2. The restoring force is maximum there, so the reason is false.

Teacher's Note:
1. Rest does not mean zero force.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( E \) = ½kA² does not depend on position, so the graph is a horizontal line.
2. The reason gives exactly this.

Teacher's Note:
1. \( E \) is constant.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( E = \frac{1}{2}m\omega^{2}A^{2} ∝ A^{2} \).
2. Doubling \( A \) quadruples \( E \), as the reason says.

Teacher's Note:
1. \( E ∝ A^{2} \).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Amplitude decay means the total energy decreases.
2. The reason names the usual causes, so it explains the assertion.

Teacher's Note:
1. Damping dissipates energy.

Answer: (c) forced oscillations

Explanation:
1. Oscillations at the frequency of an external periodic force are forced oscillations.

Teacher's Note:
1. Driven means forced.

Answer: (b) \( \text{simple harmonic with amplitude }\sqrt{a^2+b^2} \)

Explanation:
1. The two are 90° out of phase with the same ω.
2. The resultant is SHM of amplitude \( \sqrt{a^{2} + b^{2}} \).

Teacher's Note:
1. Perpendicular phasors.