A loaded vertical spring executes S.H.M. with | vedclass

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(A) For a particle executing $S.H.M.$ with time period $T$,the kinetic energy $(K)$ and potential energy $(U)$ are given by:$K = \frac{1}{2} k A^2 \co

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(A) For a particle executing $S.H.M.$ with time period $T$,the kinetic energy $(K)$ and potential energy $(U)$ are given by:$K = \frac{1}{2} k A^2 \cos^2(\omega t)$$U = \frac{1}{2} k A^2 \sin^2(\omega t)$The difference between kinetic and potential energy is:$K - U = \frac{1}{2} k A^2 (\cos^2(\omega t) - \sin^2(\omega t)) = \frac{1}{2} k A^2 \cos(2\omega t)$Since the angular frequency of the difference is $2\omega$,the new time period $T'$ is given by:$T' = \frac{2\pi}{2\omega} = \frac{T}{2}$Given $T = 4\; sec$,we have $T' = \frac{4}{2} = 2\; sec$.

$A$ loaded vertical spring executes $S.H.M.$ with a time period of $4\; sec$. The difference between the kinetic energy and potential energy of this system varies with a period of ........ $sec$.

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