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(C) The total energy $(TE)$ of a simple harmonic oscillator is constant and is given by $TE = \frac{1}{2} k A^2$,where $k$ is the force constant and $

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$1/6 \ s$

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(C) The total energy $(TE)$ of a simple harmonic oscillator is constant and is given by $TE = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.Given that the kinetic energy $(KE)$ is $75 \%$ of the total energy $(TE)$,we have $KE = 0.75 \ TE$.Since $TE = KE + PE$,the potential energy $(PE)$ must be $PE = TE - KE = TE - 0.75 \ TE = 0.25 \ TE$.Substituting the expressions for $PE$ and $TE$: $\frac{1}{2} k x^2 = 0.25 	imes (\frac{1}{2} k A^2)$.This simplifies to $x^2 = 0.25 \ A^2$,which gives $x = \pm \frac{A}{2}$.For a body starting from the origin ($x=0$ at $t=0$),the displacement is given by $x = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.Setting $x = \frac{A}{2}$,we get $\frac{A}{2} = A \sin(\omega t) \Rightarrow \sin(\omega t) = \frac{1}{2}$.This implies $\omega t = \frac{\pi}{6}$.Substituting $\omega = \frac{2\pi}{T}$,we get $(\frac{2\pi}{T}) t = \frac{\pi}{6}$.Given $T = 2 \ s$,we have $(\frac{2\pi}{2}) t = \frac{\pi}{6} \Rightarrow \pi t = \frac{\pi}{6} \Rightarrow t = \frac{1}{6} \ s$.

Column-$I$	Column-$II$
$(a)$ $y = \frac{A}{\sqrt{2}}$	$(i)$ $K = \frac{3E}{4}$
$(b)$ $y = \frac{\sqrt{3}A}{2}$	$(ii)$ $K = \frac{E}{4}$
	$(iii)$ $K = \frac{E}{2}$

Starting from the origin,a body 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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