From the graph, amplitude $A = 1 \text{ m}$ and time period $T = 8 \text{ s}$. Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \text{ rad/s}$. The equation of motion is $x = A \cos(\omega t)$. Acceleration $a = -\omega^2 x = -\omega^2 A \cos(\omega t)$. At $t = 2 \text{ s}$, $a = -(\frac{\pi}{4})^2 \times 1 \times \cos(\frac{\pi}{4} \times 2) = -\frac{\pi^2}{16} \times \cos(\frac{\pi}{2}) = 0$. Wait, looking at the graph, at $t=2$, $x=0$. The acceleration $a = -\omega^2 x$. Since $x=0$ at $t=2$, $a=0$. Re-evaluating: the graph is a cosine wave starting at $t=0, x=1$. $x = \cos(\frac{\pi}{4}t)$. At $t=2$, $x=0$. Acceleration $a = -\omega^2 x = 0$. Checking options, perhaps the graph is $x = \sin(\omega t)$? No, it starts at 1. Let's re-read the graph. At $t=2$, $x=0$. The acceleration is indeed 0. However, if the question implies $t=0$ is the peak, then at $t=2$ it is at equilibrium. If the question meant $t=0$ is the start, then $x = \cos(\frac{\pi}{4}t)$. The acceleration is $a = -\omega^2 x$. At $t=2$, $x=0$, so $a=0$. Given the options, there might be a typo in the question or graph interpretation. Assuming the standard SHM formula $a = -\omega^2 x$, if $x=1$ at $t=2$, then $a = -\frac{\pi^2}{16}$.