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(D) The total energy $(T.E.)$ of a particle in $S.H.M.$ is $T.E. = \frac{1}{2} kA^2$.The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} k(A^2

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$0.5$

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(D) The total energy $(T.E.)$ of a particle in $S.H.M.$ is $T.E. = \frac{1}{2} kA^2$.The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} k(A^2 - x^2)$.Given that $K.E. = 50 \%$ of $T.E.$,which means $K.E. = \frac{1}{2} T.E.$Substituting the expressions:$\frac{1}{2} k(A^2 - x^2) = \frac{1}{2} (\frac{1}{2} kA^2)$$A^2 - x^2 = \frac{1}{2} A^2$$x^2 = \frac{A^2}{2} \implies x = \frac{A}{\sqrt{2}}$The displacement equation for a particle starting from the mean position is $x = A \sin(\frac{2\pi t}{T})$.Substituting $x = \frac{A}{\sqrt{2}}$ and $T = 4 \ s$:$\frac{A}{\sqrt{2}} = A \sin(\frac{2\pi t}{4})$$\frac{1}{\sqrt{2}} = \sin(\frac{\pi t}{2})$$\frac{\pi t}{2} = \frac{\pi}{4}$$t = 0.5 \ s$.

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $4 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)?

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