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?

d

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2 d

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3 d

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\frac{1}{2} d

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Solution

The correct option is B $2 d$
In equilibrium $k d = m g$ or $d = \frac{m g}{k}$
In allowed to fall suddenly. if does not stop in its equilibrium position. In that case.
decrease in gravitaional PE = increase in elastic PE
or $mgd' = \frac{1}{2} k d^{2}$
or $d^{'} = \frac{2 m g}{k} = 2 d$

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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?

The correct option is B 2dIn equilibrium kd=mg or d=mgk In allowed to fall suddenly. if does not stop in its equilibrium position. In that case. decrease in gravitaional PE = increase in elastic PE or mgd'=12kd2 or d′=2mgk=2d

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?

The correct option is B $2 d$
In equilibrium $k d = m g$ or $d = \frac{m g}{k}$
In allowed to fall suddenly. if does not stop in its equilibrium position. In that case.
decrease in gravitaional PE = increase in elastic PE
or $mgd' = \frac{1}{2} k d^{2}$
or $d^{'} = \frac{2 m g}{k} = 2 d$