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A spring is hanging vertically with a mass attached to it . Gradually lowered mass stretches the spring by $x$ m . If same body attached to the same spring is released suddenly, the maximum stretch in this will be :

(A) same

(B) double

(C) triple

(D) half

My Approach - For the mass to be in equilibrium the equation will be $mg=kx$ Now I could not understand the difference in the lengths will be if we stretch it gradually or suddenly...

Please help

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The equilibrium you are talking about is static equilibrium which is not just about the net force on the mass being zero.

At the top the mass will have gravitational potential energy.
If you release the mass and assume that there are no frictional forces, the will fall losing gravitational potential energy and gaining kinetic energy as well as the spring gaining elastic potential energy.

What will happen at the static equilibrium position; will the mass suddenly stop?
N, the mass has kinetic energy and so the spring will be extended beyond the static equilibrium position.
You need to find the extension of the spring when the mass finally stops.

You know that $mg = kx$ for static equilibrium so now equate the loss in gravitational potential energy of the mass to the gain in elastic potential energy of the spring when the mass finally comes to rest.

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    $\begingroup$ Means when it is stretched suddenly kinetic energy will come and after solving by work energy theorem the answer is coming 2x . Hence the answer will be (B) double. $\endgroup$ – saladi Sep 18 '16 at 11:40
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If you let the mass fall suddenly then it will oscillate between 2 points - the point of release and the maximum stretch. If there is some air resistance it will eventually lose all of its energy and come to rest at the equilibrium point. If you lower the mass gradually it loses all of its energy immediately and stops at the equilibrium point, half way between the max. and min. points of oscillation.

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The answer will be "Double". The gradually lowered equilibrium position $(mg/k)$ will become the mean position for suddenly released case and the amplitude will become $mg/k$. Maximum stretch will be $2mg/k$.

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