1. Use the SHM potential energy relation

For a spring-mass system executing SHM, $U = \frac{1}{2}kx^2$ where $k$ is the spring constant.

From the given potential energy curve, we read the spring constant as follows:

• The graph of $U$ vs $x$ is a parabola.
• From the marked point on the graph, the potential energy is $20\,\text{J}$ at displacement $x=2\,\text{m}$.

So, $20 = \frac{1}{2}k(2)^2$ $20 = 2k$ $k = 10\,\text{N/m}$

2. Find the time period of the spring-mass system

Given mass, $m = 5\,\text{kg}$

Time period of spring-mass SHM: $T = 2\pi\sqrt{\frac{m}{k}}$

Substitute values: $T = 2\pi\sqrt{\frac{5}{10}} = 2\pi\sqrt{\frac{1}{2}}$

3. Use the pendulum time period

For a simple pendulum, $T = 2\pi\sqrt{\frac{L}{g}}$

Given length, $L = 4\,\text{m}$

Since both periods are equal, $2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{L}{g}}$

So, $\frac{m}{k} = \frac{L}{g}$

Hence, $g = \frac{Lk}{m}$

Substitute values: $g = \frac{4\times 10}{5} = 8\,\text{m/s}^2$

4. Compare with options

The obtained value is $g=8\,\text{m/s}^2$

This does not match any of the given options.

Therefore, the stored correct answer $\text{C} = 4\,\text{m/s}^2$ is inconsistent with the standard SHM calculation based on the graph reading $U=20\,\text{J}$ at $x=2\,\text{m}$.

If instead the graph indicates a different point, the answer may change, but with the usual reading above, the correct value should be $8\,\text{m/s}^2$.