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 x. 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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Solution

The correct option is B

When spring is gradually lowered to it's equilibrium position

$k x = m g ∴ x = \frac{m g}{k} .$

When spring is allowed to fall suddenly it oscillates about it's mean position

Let y is the maximum elongation, then at lower extreme, by the conservation of energy

$\Rightarrow \frac{1}{2} k y^{2} = m g y \Rightarrow y = \frac{2 m g}{k} = 2 x .$

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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 x. 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 When spring is gradually lowered to it's equilibrium position kx=mg∴x=mgk. When spring is allowed to fall suddenly it oscillates about it's mean position Let y is the maximum elongation, then at lower extreme, by the conservation of energy ⇒12ky2=mgy⇒y=2mgk=2x.