The x - t graph of a particle performing simple harmonic motion is...

From the figure, it can be concluded that

$\begin{cases}A & = & 1 \\ \omega & = & \dfrac{2\pi}{8} \\ = & \dfrac{\pi}{4}\end{cases}$

Hence, the equation of motion of the particle is given by

$\begin{cases}x & = & Asin\omegat \\ = & sin\dfrac{\pi}{4}t . . . \left(1\right)\end{cases}$

The velocity of the particle is, then, given by

$\begin{cases}v & = & \dfrac{dx}{dt} \\ = & \dfrac{d}{dt}\left(sin\dfrac{\pi}{4}t\right) \\ = & \dfrac{\pi}{4}cos\dfrac{\pi}{4}t . . . \left(2\right)\end{cases}$

Hence, the acceleration of the particle can be written as

$\begin{cases}a & = & \dfrac{dv}{dt} \\ = & \dfrac{d}{dt}\left(\dfrac{\pi}{4}cos\dfrac{\pi}{4}t\right) \\ = & - \dfrac{{\pi}^{2}}{16}sin\dfrac{\pi}{4}t . . . \left(3\right)\end{cases}$

Substitute $2$ for $t$ into equation (3) to calculate the required value of the acceleration.

$\begin{cases}a & = & - \dfrac{{\pi}^{2}}{16}\sin \left(\dfrac{\pi}{4} \times 2\right) \\ = & - \dfrac{{\pi}^{2}}{16}{ms}^{- 2}\end{cases}$