Animated Solution for Physics - Oscillations: The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy $U(x)$ versus time ($t$) plot of the particle is correctly shown in figure :

Imagine a mass attached to a spring, oscillating back and forth on a frictionless surface. The graph provided in the question captures this exact motion, plotting the displacement $x$ of the particle against time $t$.

Notice how the curve starts right at the origin $(0,0)$, rises to a positive peak, and then gracefully crosses the time axis at points $A$, $B$, and $C$. This is the classic, unmistakable signature of a sine wave. Because it starts from zero, we can confidently write the equation of motion as:

Here, $x_0$ is the maximum displacement (the amplitude), and $\omega$ is the angular frequency. At the specific points $O$, $A$, $B$, and $C$, the particle is passing through its mean position, meaning its displacement is exactly zero.

Now, let's shift our focus to the energy of this system. For a particle executing simple harmonic motion, the restoring force is given by Hooke's Law, $F=-kx$. The potential energy $U$ stored in the system (like the energy stored in a compressed or stretched spring) depends purely on its position. The fundamental formula is:

To see how this energy evolves over time, we need to merge our two equations. Let's carefully substitute our expression for $x(t)$ into the potential energy formula:

When we square the terms inside the bracket, something beautiful happens. We get:

That constant part in the parentheses, $\frac{1}{2}k{x}_0^2$, is simply the maximum potential energy the system can hold, which we can call $U_{max}$. So, our function elegantly simplifies to:

There is a crucial physical and mathematical catch here. Because of the square, the ${\sin }^2(\omega t)$ function can never be negative. Think about it physically: whether the spring is stretched to the right or compressed to the left, it stores positive energy. The potential energy is always greater than or equal to zero.

Furthermore, whenever the displacement is zero—at points $O$, $A$, $B$, and $C$—the potential energy must also drop exactly to zero, because the spring is completely relaxed at the mean position.

Armed with this knowledge, let's evaluate the given options.

Graphs (a) and (b) show the potential energy dipping into negative territory, which violates the fundamental physics of a spring system. Graph (c) starts at a maximum energy at $t=0$, which contradicts our finding that the energy must be zero when the displacement is zero at the start.

Only Graph (d) stays entirely positive and correctly touches zero at the exact moments the particle passes through the mean position ($O$, $A$, $B$, and $C$). It perfectly matches our derived ${\sin }^2(\omega t)$ function, making it the undeniably correct answer.