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:

We need to identify the correct potential energy ($$U$$) versus time ($$t$$) graph for a particle executing free simple harmonic motion (SHM), given its displacement-time curve.

1. Physics Analysis

• Initial Position ($t = 0$): The given displacement graph shows that at $$t = 0$$, the particle starts from the mean position ($$x = 0$$).

  Therefore, its kinematic equation is: $$x(t) = A \sin(\omega t)$$.

• Potential Energy Equation: The potential energy of a simple harmonic oscillator is given by:

  $$U(t) = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2(\omega t)$$

2. Key Graphical Rules

1. Starts at Zero: At $$t = 0$$, $$\sin^2(0) = 0$$, so the graph must begin precisely at the origin $$(0,0)$$.
2. Strictly Non-Negative: Because the sine function is squared ($$\sin^2(\omega t)$$), the potential energy value can never drop below zero ($$U \ge 0$$). The entire curve must lie completely on or above the time axis.

3. Matching with Options

Looking at the option:

• The option D represents a series of entirely positive, upward loops starting at the origin.

Final Answer: Option D (The plot containing entirely positive, upward-only loops starting from the origin)