A spring of force constant 200 Nm-1 has a block of mass 10 kg hanging at its one end and the other end of the spring is attached to the ceiling of an elevator. The elevator is rising upwards with an acceleration of π4 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? (round off to nearest integer)

When elevator is moving up, using NLM,T−mg=mg4⇒5mg4This tension elongates the spring by x,T=kx⇒5mg4kThe equilibrium position of the block is, x0=mgk So the amplitude is, A=x-x0=5mg4k-mgk=mg4k=0.125m

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A spring of force constant $200 {Nm}^{- 1}$ has a block of mass 10 kg hanging at its one end and the other end of the spring is attached to the ceiling of an elevator. The elevator is rising upwards with an acceleration of $\frac{π}{4}$ 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? (round off to nearest integer)

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Detailed Solution

When elevator is moving up, using NLM,
$T - m g = \frac{m g}{4} \Rightarrow \frac{5 m g}{4}$
This tension elongates the spring by x,
$T = kx \Rightarrow \frac{5 m g}{4 k}$
The equilibrium position of the block is, $x_{0} = \frac{m g}{k}$
So the amplitude is,
$A = x - x_{0} = \frac{5 mg}{4 k} - \frac{mg}{k} = \frac{mg}{4 k} = 0.125 m$