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A rigid spring of spring constant k is fixed to the ceiling of a lift. The other end of the spring is attached to a block of mass m. The block hangs in equilibrium. Now the lift starts accelerating downwards with an acceleration 2g. [Assume the length of the spring to be large and its mass is negligible]. Now,

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the block would not perform simple harmonic motion and would stick to the ceiling

the block performs SHM with time period

T = 2 π \sqrt{\frac{m}{k}}

the block performs oscillatory motion but not SHM

the block performs simple harmonic motion with amplitude

\frac{2mg}{k}

answer is D, B.

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Detailed Solution

Time period of spring mass system independent of acceleration of lift

$∴ T = 2 π \sqrt{\frac{m}{k}}$

When the lift were at rest at equilibrium $k x_{0} = mg$

When the left moves down ward with an acceleration 2g, block experience pseudeo force 4 g upward

Question Image

$k x_{0} = mg \Rightarrow 2 mg = kA - k x_{0} + mg$

$A = \frac{2mg}{k}$

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