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A spring mass system is hanging from the ceiling of an elevator which is at rest. The elevator starts accelerating upwards with acceleration a. Its motion is observed with respect to the elevator

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Time period of oscillation is

2 π 2 m k

Time period of oscillation is

2 π m k

Amplitude of oscillation is ma/2k

Amplitude of oscillation is m(g + a) /2k

answer is C.

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Detailed Solution

When elevator is at rest

m g = k x 0 − − − (1)

Apply psendo force of ma. The mean position shifts downwards such that the distance between initial and new equilibrium positions is amplitude of oscillation A.

k x 0 + A = m g + m a

∴ A = m a k

Net restoring force when the block is x units below the mean position

= − k x 0 + A + x + m g + m a = − k x

∴ − k x = m a = m ω 2 x

∴ ω = k m

∴ τ = 2 π ω = 2 π m k

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