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(C) The displacement of a particle in $SHM$ is given by $x(t) = A \cos(\omega t + \phi)$. From the figure,at $t = 0$,$x$ is at its maximum positive va

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(C) The displacement of a particle in $SHM$ is given by $x(t) = A \cos(\omega t + \phi)$. From the figure,at $t = 0$,$x$ is at its maximum positive value,so $x(t) = A \cos(\omega t)$.The potential energy $(PE)$ of a particle in $SHM$ is given by $PE = \frac{1}{2} k x^2$.Substituting $x(t) = A \cos(\omega t)$ into the $PE$ equation:$PE = \frac{1}{2} k (A \cos(\omega t))^2 = \frac{1}{2} k A^2 \cos^2(\omega t)$.Using the trigonometric identity $\cos^2(	heta) = \frac{1 + \cos(2	heta)}{2}$:$PE = \frac{1}{2} k A^2 \left( \frac{1 + \cos(2\omega t)}{2} ight) = \frac{1}{4} k A^2 (1 + \cos(2\omega t))$.At $t = 0$,$PE = \frac{1}{2} k A^2$,which is the maximum value.As $t$ increases,$\cos(2\omega t)$ decreases,so $PE$ decreases to $0$ when $2\omega t = \pi/2$ (or $t = \pi/4\omega$),and then increases back to the maximum value when $2\omega t = \pi$ (or $t = \pi/2\omega$).This behavior matches the graph in option $C$.

The displacement-time graph of a particle executing $SHM$ is as shown in the figure. The corresponding graph between potential energy $(PE)$ and time is:

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