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(D) The potential energy $(PE)$ of a particle of mass $m$ executing simple harmonic motion is given by the formula $PE = \frac{1}{2} m \omega^2 y^2$,w

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(D) The potential energy $(PE)$ of a particle of mass $m$ executing simple harmonic motion is given by the formula $PE = \frac{1}{2} m \omega^2 y^2$,where $y$ is the displacement from the equilibrium position $O$.At the equilibrium position $O$,the displacement $y = 0$,so the potential energy $PE = 0$.At the extreme points $X_1$ and $X_2$,the displacement $y$ is equal to the amplitude $A$ (i.e.,$y = \pm A$).Thus,the potential energy at both extreme points is $PE = \frac{1}{2} m \omega^2 A^2$,which is the maximum value.The relationship $PE \propto y^2$ represents a parabola opening upwards,with its vertex at the origin $(0,0)$.Therefore,the graph of potential energy versus displacement is a parabola,which corresponds to the graph shown in option $D$.

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

$A$ particle of mass $m$ oscillates with simple harmonic motion between points $X_1$ and $X_2$,the equilibrium position being $O$. Its potential energy will be as shown in the following graph:

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