To solve this problem, we need to determine the acceleration of the particle at $t = 2s$ from the given $x-t$ graph.$\\$The equation for simple harmonic motion (SHM) is:$\\ x(t) = A \cos(\omega t + \phi) \\$Where:$\\$• $A$ is the amplitude.$\\$• $\omega$ is the angular frequency.$\\$• $\phi$ is the phase constant.$\\$From the graph:$\\$• The amplitude $A = 1$ m.$\\$• The period $T = 8$ s (since the graph completes one cycle in 8 s).$\\$The angular frequency $\omega$ is given by:$\\ \omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4}$ rad/s.$\\$The acceleration in SHM is given by:$\\ a(t) = -\omega^2 x(t) \\$At $t = 2s,$ from the graph, $x(2) = 1$ m.$\\$Substitute the values into the acceleration equation:$\\ a(2) = -\left(\frac{\pi}{4}\right)^2 \cdot 1 \\ a(2) = -\frac{\pi^2}{16}$ m/s$^2 \\$Therefore, the acceleration of the particle at $t = 2s$ is $-\frac{\pi^2}{16}$ m/s$^2. \\$This corresponds to Option 4.