A body is attached to the lower end of a vertical helical 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?

x

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2 x

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3 x

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\frac{x}{2}

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Solution

The correct option is B. $2 x$ .
When body is lowered gradually, its weight acts at C.G. of the spring. When same body is allowed to fall freely, the same weight acts at lower end of the spring. In the latter case, original length (L) of spring is double.
As $Δ L \propto L$ , therefore, $Δ L$ becomes twice in second case i.e. $2 x$ .

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A body is attached to the lower end of a vertical helical 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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