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(A) The total energy $E$ of a simple harmonic oscillator is given by $E = \frac{1}{2} m \omega^2 A^2$.The kinetic energy $KE$ at any time $t$ is given

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

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(A) The total energy $E$ of a simple harmonic oscillator is given by $E = \frac{1}{2} m \omega^2 A^2$.The kinetic energy $KE$ at any time $t$ is given by $KE = E \cos^2(\omega t)$ (since the body starts from the mean position,the displacement $x = A \sin(\omega t)$,so velocity $v = A \omega \cos(\omega t)$).Given that $KE = 75\% 	ext{ of } E$,we have:$KE = \frac{75}{100} E = \frac{3}{4} E$.Substituting the expression for $KE$:$E \cos^2(\omega t) = \frac{3}{4} E$$\cos^2(\omega t) = \frac{3}{4}$$\cos(\omega t) = \frac{\sqrt{3}}{2}$.We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,so $\omega t = \frac{\pi}{6}$.Given the period $T = 2\,s$,the angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi 	ext{ rad/s}$.Substituting $\omega$ into the equation:$\pi t = \frac{\pi}{6}$$t = \frac{1}{6}\,s$.

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

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