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(D) The total energy of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.Since the particle starts from the origin ($x=0$ a

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$\frac{1}{6} \ s$

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(D) The total energy of a particle performing $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.Since the particle starts from the origin ($x=0$ at $t=0$),its displacement is $x = A \sin(\omega t)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.The kinetic energy is $K = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.Given that $K = 75\% \ E = \frac{3}{4} E$,we have:$\frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = \frac{3}{4} (\frac{1}{2} m A^2 \omega^2)$.$\cos^2(\omega t) = \frac{3}{4} \Rightarrow \cos(\omega t) = \frac{\sqrt{3}}{2}$.Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have $\omega t = \frac{\pi}{6}$.Substituting $\omega = \frac{2 \pi}{T}$,we get $(\frac{2 \pi}{T}) t = \frac{\pi}{6}$.$t = \frac{T}{12}$.Given $T = 2 \ s$,$t = \frac{2}{12} = \frac{1}{6} \ s$.

Starting from the origin,a particle oscillates simple harmonically with a time period of $2 \ s$. After what time will its kinetic energy be $75 \%$ of the total energy?

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