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A body is attached to the lower end of a vertical spiral spring and it is gradually lowered to its equilibrium position. This stretches the spring by a length d. If the same body attached to the same spring is allowed to fall suddenly, what would be the maximum stretching in this case?

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(1/2)d

answer is B.

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Detailed Solution

In equilibrium, $k d = m g or d = \frac{m g}{k}$ ... (i)

If allowed to fall suddenly, the body does not stop in its equilibrium position. In that case,

decrease in gravitational PE = increase in elastic PE

or $m g d^{'} = \frac{1}{2} k d^{' 2} \Rightarrow d^{'} = \frac{2 m g}{k} = 2 d$ [From Eq. (i)]

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