A particle of mass $10\, gm$ is describing $S.H.M.$ along a straight line with period of $2\, sec$ and amplitude of $10, cm$. Its kinetic energy when it is at $5 \,cm$ from its equilibrium position is
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(c) Kinetic energy $K = \frac{1}{2}m{\omega ^2}({a^2} - {y^2})$
$ = \frac{1}{2} \times 10 \times {\left( {\frac{{2\pi }}{2}} \right)^2}[{10^2} - {5^2}]$

$ = 375\;{\pi ^2}ergs$

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