JEE Physics Oscillations MCQs Set 05

Answer: (b) amplitude

Explanation:
1. \( E = \dfrac{1}{2}m\omega^2A^2 \), which is proportional to \( A^2 \).
2. It is directly proportional to the square of the amplitude (and to m, but only mass to the first power).

Teacher's Note:
1. Total energy also varies as 1/T².

Answer: (d) E

Explanation:
1. \( E = \dfrac{1}{2}m\omega^2A^2 = \dfrac{2\pi^2mA^2}{T^2} \).
2. With A and T both doubled, the factor \( \dfrac{A^2}{T^2} \) is unchanged.
3. The energy remains E.

Teacher's Note:
1. Energy is proportional to A²/T².

Answer: (d) \( 3E/4 \)

Explanation:
1. \( PE = \dfrac{1}{2}m\omega^2x^2 = E\dfrac{x^2}{A^2} = \dfrac{E}{4} \) at \( x = \dfrac{A}{2} \).
2. \( KE = E - \dfrac{E}{4} = \dfrac{3E}{4} \).

Teacher's Note:
1. KE = E(1 − x²/A²).

Answer: (b) \( E/4 \)

Explanation:
1. \( PE = E\left(\dfrac{x}{A}\right)^2 = E\times\dfrac{1}{4} = \dfrac{E}{4} \).

Teacher's Note:
1. Potential energy goes as x².

Answer: (c) \( 2\sqrt{2} \) cm

Explanation:
1. Amplitude \( A = 4 \) cm.
2. KE = PE means \( \dfrac{1}{2}m\omega^2x^2 = \dfrac{1}{2}\cdot\dfrac{1}{2}m\omega^2A^2 \).
3. \( x = \dfrac{A}{\sqrt{2}} = \dfrac{4}{\sqrt{2}} = 2\sqrt{2} \) cm.

Teacher's Note:
1. At this point each energy is E/2.

Answer: (b) \( \dfrac{1}{4}kA^2 \)

Explanation:
1. The average of \( \sin^2 \) or \( \cos^2 \) over a cycle is 1/2.
2. \( \langle PE\rangle = \dfrac{1}{2}\times\dfrac{1}{2}kA^2 = \dfrac{1}{4}kA^2 \).

Teacher's Note:
1. Average KE equals average PE, each half of the total energy.

Answer: (c) 1000 N/m

Explanation:
1. Maximum KE equals the total energy \( \dfrac{1}{2}kA^2 \).
2. \( k = \dfrac{2\times5}{(0.1)^2} = 1000 \) N/m.

Teacher's Note:
1. Force constant is measured in N/m.

Answer: (d) 32 dyne

Explanation:
1. \( F = -\dfrac{dU}{dx} = -16x \).
2. At \( x = -2 \): \( F = +32 \) dyne (directed towards the origin).

Teacher's Note:
1. The mass is not needed to find the force.

Answer: (c) \( \dfrac{a}{2} \)

Explanation:
1. \( \dfrac{1}{2}m\omega^2x^2 = \dfrac{1}{4}\cdot\dfrac{1}{2}m\omega^2a^2 \).
2. \( x^2 = \dfrac{a^2}{4} \), so \( x = \dfrac{a}{2} \).

Teacher's Note:
1. Potential energy depends on the square of the displacement.

Answer: (c) \( \dfrac{4}{3} \)

Explanation:
1. At the mean position \( KE = E \).
2. At half amplitude \( KE = \dfrac{3E}{4} \).
3. Ratio \( = \dfrac{E}{3E/4} = \dfrac{4}{3} \).

Teacher's Note:
1. Total energy E is the maximum kinetic energy.

Answer: (c) \( \dfrac{1}{\sqrt{2}} \)

Explanation:
1. \( \dfrac{1}{2}m\omega^2x^2 = \dfrac{1}{2}m\omega^2(A^2 - x^2) \Rightarrow 2x^2 = A^2 \).
2. \( \dfrac{x}{A} = \dfrac{1}{\sqrt{2}} \).

Teacher's Note:
1. Each is half of the total energy.

Answer: (c) \( 9U \)

Explanation:
1. \( U = \dfrac{1}{2}kx^2 \propto x^2 \).
2. Elongation is three times, so the energy is 9U.

Teacher's Note:
1. Potential energy varies as the square of extension.

Answer: (c) \( \dfrac{E}{4} \)

Explanation:
1. \( U \propto x^2 \), so for \( \dfrac{x}{2} \) the energy is \( \dfrac{E}{4} \).

Teacher's Note:
1. Half the extension gives a quarter of the energy.

Answer: (d) \( f\sqrt{2} \)

Explanation:
1. The two springs act in parallel, so the effective constant is \( 2K \).
2. \( f' = \dfrac{1}{2\pi}\sqrt{\dfrac{2K}{M}} = \sqrt{2}\,f \).

Teacher's Note:
1. Springs on opposite sides of the mass both push or pull in the same direction.

Answer: (c) \( \sqrt{K_1 + K_2} \)

Explanation:
1. The restoring forces add, so \( K_{eff} = K_1 + K_2 \).
2. \( f = \dfrac{1}{2\pi}\sqrt{\dfrac{K_1+K_2}{M}} \propto \sqrt{K_1+K_2} \).

Teacher's Note:
1. Parallel springs: constants add.

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

Explanation:
1. \( \nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \propto \dfrac{1}{\sqrt{m}} \).
2. Mass four times gives frequency \( \dfrac{\nu}{2} \).

Teacher's Note:
1. Heavier mass, lower frequency.

Answer: (d) \( \dfrac{K_1K_2}{K_1 + K_2} \)

Explanation:
1. In series the extensions add: \( \dfrac{1}{K_{eff}} = \dfrac{1}{K_1} + \dfrac{1}{K_2} \).
2. \( K_{eff} = \dfrac{K_1K_2}{K_1+K_2} \).

Teacher's Note:
1. Springs in series are softer than either spring.

Answer: (d) \( 2\pi\sqrt{\dfrac{M}{K_1 + K_2}} \)

Explanation:
1. Parallel springs share the load, so \( K_{eff} = K_1 + K_2 \).
2. \( T = 2\pi\sqrt{\dfrac{M}{K_1+K_2}} \).

Teacher's Note:
1. Parallel arrangement is stiffer.

Answer: (b) \( \left[\dfrac{1}{2K_1} + \dfrac{1}{K_2}\right]^{-1} \)

Explanation:
1. The two upper springs are in parallel: \( 2K_1 \).
2. This combination is in series with \( K_2 \).
3. \( \dfrac{1}{K_{eff}} = \dfrac{1}{2K_1} + \dfrac{1}{K_2} \), so \( K_{eff} = \left[\dfrac{1}{2K_1} + \dfrac{1}{K_2}\right]^{-1} \).

Teacher's Note:
1. Add parallel constants first, then combine in series using reciprocals.

Answer: (d) 1.6 kg

Explanation:
1. \( T^2 \propto m \): \( \dfrac{2^2}{m} = \dfrac{3^2}{m+2} \).
2. \( 4(m+2) = 9m \), so \( m = 1.6 \) kg.

Teacher's Note:
1. Square the periods before taking the ratio.