Acceleration of Particle from Displacement Time Graph in SHM

From the given $x$-$t$ graph, the particle undergoes simple harmonic motion (SHM).

1. Amplitude and Time Period:

   • The maximum displacement (amplitude) of the particle is:
     $A = 1\text{ cm}$
   • The particle starts at $x = 0$ with a positive velocity at $t = 0\text{ s}$ and completes one full oscillation at $t = 8\text{ s}$. Thus, the time period of oscillation is:
     $T = 8\text{ s}$

2. Angular Frequency:
   The angular frequency $\omega$ is given by:
   $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4}\text{ rad/s}$

3. Equation of Motion:
   Since $x(0) = 0$ and the particle is moving in the positive direction initially, the displacement as a function of time can be written as:
   $x(t) = A \sin(\omega t) = \sin\left(\frac{\pi}{4} t\right)\text{ cm}$

4. Displacement at $t = \frac{4}{3}\text{ s}$:
   $x\left(\frac{4}{3}\right) = \sin\left(\frac{\pi}{4} \times \frac{4}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\text{ cm}$

5. Acceleration of the Particle:
   The acceleration $a(t)$ in simple harmonic motion is related to the displacement by:
   $a(t) = -\omega^2 x(t)$

   Substituting the values at $t = \frac{4}{3}\text{ s}$:
   $a = -\left(\frac{\pi}{4}\right)^2 \times \left(\frac{\sqrt{3}}{2}\right) = -\frac{\pi^2}{16} \times \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{32}\pi^2\text{ cm/s}^2$

Therefore, the correct option is D.