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A spring of force constant 200Nm^(-1) ha...

A spring of force constant `200Nm^(-1)` has a block of mass 10 kg hanging at its one end and the other end of the spring is attached to the celling of an elevator. The elevator is rising upwards with an acceleration of `g/4` and the block is in equilibrium with respect to the elevator . when the acceleration of the elevator suddenly ceases , the block starts oscillating . What is the amplitude (in m) of these oscillations ?

Text Solution

Verified by Experts

The correct Answer is:
0.125
When elevator is moving up,
`T-mg=(mg)/(4)impliesT=(5mg)/(4)`
This tension elongates the spring by X,
`T=kximpliesx=(5mg)/(4k)`
The equilibrium position of the block is,
`x_(0)=(mg)/(k)`
So the amplitude is,
`A=x-x_(0)=(5mg)/(4k)-(mg)/(k)=(mg)/(4k)=0.125m`
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