1. For a particle in SHM, the potential energy is $U= frac12 kx^2= frac12 m\omega^2 x^2.$ So the potential energy depends on the square of displacement.

2. From the given $x$ vs $t$ graph, the displacement varies sinusoidally in time. If $x=A\sin(\omega t+\phi),$ then $U \propto x^2=A^2\sin^2(\omega t+\phi).$

3. Therefore, the important features of the $U$ vs $t$ graph are:

• $U$ is never negative.
• $U$ is maximum when $|x|=A$.
• $U=0$ when $x=0$.
• Since $\sin^2\theta$ has period $\pi$, the period of $U$ is $T_U=\frac{T}{2},$ where $T$ is the period of SHM.

4. So the correct graph must:

• stay entirely above or on the time axis,
• touch zero whenever displacement crosses zero,
• have maxima at both positive and negative extreme displacements,
• oscillate with double frequency compared to $x(t)$.

5. Among the given options, the graph satisfying these conditions is Option D.

Hence, the correct answer is: $\boxed{D}$