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A block of m is suspended by a spring and deflection of the spring is found to be A. Now the block is pulled downward by a distance A and released. Then energy of oscillation the block is

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$\frac{1}{4} mgA$

$2 mgA$

$\frac{1}{2} mgA$

$mgA$

answer is C.

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Detailed Solution

When the block is in equilibrium

mg = KA

After release the block will oscillate with amplitude A.

Angular frequency of the block, $w = \sqrt{\frac{k}{m}}$

Energy of oscillation $= \frac{1}{2} {mw}^{2} A^{2}$

$= \frac{1}{2} m \frac{k}{m} A^{2} = \frac{1}{2} {kA}^{2} = \frac{1}{2} mgA$

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