JEE Physics Oscillations MCQs Set 06

Answer: (a) T

Explanation:
1. For both arrangements \( T = 2\pi\sqrt{\dfrac{m}{k}} \).
2. Gravity only shifts the equilibrium position in the vertical case; it does not change the period.

Teacher's Note:
1. Period of a spring pendulum does not depend on g.

Answer: (b) \( \sqrt{2}\,\nu \)

Explanation:
1. Each half-spring has force constant \( 2K \).
2. \( \nu' = \dfrac{1}{2\pi}\sqrt{\dfrac{2K}{m}} = \sqrt{2}\,\nu \).

Teacher's Note:
1. Shorter spring, stiffer spring.

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

Explanation:
1. Energy \( U = \dfrac{F^2}{2K} \).
2. Equal energies: \( \dfrac{F_1^2}{K_1} = \dfrac{F_2^2}{K_2} \).
3. \( F_1 : F_2 = \sqrt{K_1} : \sqrt{K_2} \).

Teacher's Note:
1. Write U in terms of the force F instead of the extension.

Answer: (b) \( \dfrac{4}{3}K \)

Explanation:
1. The force constant is inversely proportional to the length.
2. The remaining length is \( \dfrac{3}{4} \), so \( K' = \dfrac{4}{3}K \).

Teacher's Note:
1. A shorter spring is stiffer.

Answer: (c) \( T/2 \)

Explanation:
1. Each quarter has force constant \( 4K \).
2. \( T' = 2\pi\sqrt{\dfrac{m}{4K}} = \dfrac{T}{2} \).

Teacher's Note:
1. Four times the stiffness halves the period.

Answer: (c) \( \dfrac{1}{2\pi}\sqrt{\dfrac{3k}{m}} \)

Explanation:
1. The restoring forces add: \( k_{eff} = k + 2k = 3k \).
2. \( f = \dfrac{1}{2\pi}\sqrt{\dfrac{3k}{m}} \).

Teacher's Note:
1. Springs on two sides of the mass act in parallel.

Answer: (c) \( \dfrac{k_1k_2}{k_1 + k_2} \)

Explanation:
1. For springs in series \( \dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2} \).
2. So \( k = \dfrac{k_1k_2}{k_1+k_2} \).

Teacher's Note:
1. Same force through both springs, extensions add.

Answer: (c) 20 s

Explanation:
1. \( T \propto \sqrt{m} \).
2. Mass four times gives period \( \sqrt{4} = 2 \) times: 20 s.

Teacher's Note:
1. Period grows as the square root of mass.

Answer: (b) 2 times

Explanation:
1. \( v_{max} = A\omega \) and ω depends only on L and g.
2. Doubling A doubles \( v_{max} \).

Teacher's Note:
1. The period is unaffected by the amplitude for small angles.

Answer: (c) 4 s

Explanation:
1. \( T = 2\pi\sqrt{\dfrac{L}{g}} = 2\pi\sqrt{\dfrac{39.2}{9.8\pi^2}} = 2\pi\times\dfrac{2}{\pi} = 4 \) s.

Teacher's Note:
1. 39.2/9.8 = 4.

Answer: (d) become twice

Explanation:
1. \( T \propto \sqrt{L} \).
2. Four times the length gives twice the period.

Teacher's Note:
1. Square-root dependence on length.

Answer: (a) 1 m

Explanation:
1. A seconds pendulum has \( T = 2 \) s.
2. \( L = \dfrac{gT^2}{4\pi^2} = \dfrac{9.8\times4}{39.5} \approx 0.99 \) m.

Teacher's Note:
1. Its half period, 1 s, is the time between extremes.

Answer: (d) infinity

Explanation:
1. In an orbiting satellite everything is in free fall, so the effective gravity on the bob is zero.
2. With g_eff = 0 there is no restoring force and the pendulum does not oscillate: the period is infinite.

Teacher's Note:
1. Weightlessness means no restoring force.

Answer: (b) 59.8 min

Explanation:
1. For large length, \( T = 2\pi\sqrt{\dfrac{1}{g\left(\dfrac{1}{R}+\dfrac{1}{L}\right)}} \).
2. With \( L = R \): \( T = 2\pi\sqrt{\dfrac{R}{2g}} = \dfrac{84.6}{\sqrt{2}} = 59.8 \) min.

Teacher's Note:
1. 84.6 min is the limit when the length tends to infinity.

Answer: (d) 1½ hours

Explanation:
1. For \( L \to \infty \), \( T = 2\pi\sqrt{\dfrac{R}{g}} = 2\pi\sqrt{\dfrac{6.4\times10^6}{9.8}} = 5077 \) s = 84.6 min.
2. That is about 1½ hours.

Teacher's Note:
1. This equals the orbital period close to the earth.

Answer: (b) \( \dfrac{\sqrt{3}\,T}{2} \)

Explanation:
1. Effective gravity \( g' = g + \dfrac{g}{3} = \dfrac{4g}{3} \).
2. \( T' = T\sqrt{\dfrac{g}{g'}} = T\sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{2}T \).

Teacher's Note:
1. Upward acceleration increases the effective g and reduces the period.

Answer: (b) decrease

Explanation:
1. Standing raises the centre of mass of the child, so the effective length of the swing decreases.
2. Since \( T \propto \sqrt{L} \), the period decreases.

Teacher's Note:
1. Compare a pendulum whose bob is moved nearer to the support.

Answer: (c) \( \sqrt{3}:2 \)

Explanation:
1. System A: \( K_A = \dfrac{K}{2} \).
2. System B: the parallel pair gives \( 2K \) in series with K: \( K_B = \dfrac{2K\times K}{3K} = \dfrac{2K}{3} \).
3. \( \dfrac{f_A}{f_B} = \sqrt{\dfrac{K_A}{K_B}} = \sqrt{\dfrac{1/2}{2/3}} = \sqrt{\dfrac{3}{4}} = \sqrt{3} : 2 \).

Teacher's Note:
1. Reduce each system to a single effective spring first.

Answer: (d) \( nK \)

Explanation:
1. Force constant is inversely proportional to length.
2. Each part has \( \dfrac{1}{n} \) of the length, so the constant is \( nK \).

Teacher's Note:
1. Reverse of connecting n springs in series.

Answer: (b) the effective value of g becomes zero in the artificial satellite

Explanation:
1. In orbit the bob is in free fall, so the effective g is zero.
2. \( T = 2\pi\sqrt{L/g_{eff}} \) becomes infinite and the pendulum does not oscillate.

Teacher's Note:
1. A spring watch still works because it does not depend on gravity.