For a particle executing SHM, the displacement $x$ is given by $x=A\cos \omega t$. Choose the graph which represents the variation of potential energy as a function of time $t$ and displacement $x$.

The correct option is B $I,IV$
The equation of SHM is given as
$x=A\cos \omega t$
The potential energy $\left(PE\right)$ of a particle performing SHM is given by $PE=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$
The potential energy of the particle should be minimum (in this case zero) at mean position $(x=0)$ and maximum at extreme position at $(x=+A,t=0)$. Hence, the graph $I$ is correct.

Further, in the graph $IV$, potential energy is minimum at $x=0$. Hence, this is also correct.
So, option $\left(b\right)$ is the correct answer.

Why this question? Tip: Potential energy is minimum at mean position and maximum at extreme position in SHM.