Step-by-step Solution:

1. Analyze the given x-t graph: The graph shows the position x of a particle as a function of time t. It represents a Simple Harmonic Motion (SHM).

   • Amplitude (A): The maximum displacement from the equilibrium position is the amplitude. From the graph, the maximum value of x is 1 cm. So, A = 1 cm.
   • Time Period (T): The time taken to complete one full oscillation is the time period. The particle starts at x = 1 cm (positive extreme) at t = 0 s and returns to x = 1 cm at t = 8 s. Thus, the time period T = 8 s.

2. Determine the angular frequency (ω): The angular frequency is related to the time period by the formula: $\omega = {{2\pi } \over T}$ Substituting T = 8 s, we get: $\omega = {{2\pi } \over 8} = {\pi \over 4} \text{ rad/s}$

3. Write the equation of motion: Since the particle is at its positive extreme position (x = +A) at t = 0, the equation for its position x(t) can be described by a cosine function with zero phase shift: $x(t) = A \cos(\omega t)$ Substituting the values of A and ω: $x(t) = 1 \cdot \cos\left({\pi \over 4}t\right) = \cos\left({\pi \over 4}t\right) \text{ cm}$

4. Find the expression for acceleration: The acceleration a(t) in SHM is the second derivative of the position with respect to time, or more simply, it is given by the relation $a(t) = -\omega^2 x(t)$. Substituting the expressions for ω and x(t): $a(t) = -\left({\pi \over 4}\right)^2 \cos\left({\pi \over 4}t\right)$ $a(t) = -{{\pi^2} \over {16}} \cos\left({\pi \over 4}t\right) \text{ cm/s}^2$

5. Calculate the acceleration at t = 4/3 s: We need to find the value of acceleration at the specific time t = 4/3 s. We substitute this value into the acceleration equation: $a\left({4 \over 3}\right) = -{{\pi^2} \over {16}} \cos\left({\pi \over 4} \cdot {4 \over 3}\right)$ $a\left({4 \over 3}\right) = -{{\pi^2} \over {16}} \cos\left({\pi \over 3}\right)$ We know that $\cos(\pi/3) = \cos(60^\circ) = 1/2$. $a\left({4 \over 3}\right) = -{{\pi^2} \over {16}} \cdot \left({1 \over 2}\right)$ $a\left({4 \over 3}\right) = -{{\pi^2} \over {32}} \text{ cm/s}^2$

6. Compare with the given options: The calculated acceleration is $-{\pi^2} / {32}$ cm/s$^2$. This matches option B.

7. Conclusion: Based on the provided graph and the question asking for acceleration at t=4/3 s, the correct answer is $-{\pi^2} / {32}$ cm/s$^2$. The stored correct answer is D, which is incorrect. Answer D, $-{{\sqrt 3 } \over {32}}{\pi ^2}$, would be the correct answer if the time was t=2/3 s, as $\cos((\pi/4)(2/3)) = \cos(\pi/6) = \sqrt{3}/2$, which suggests a possible typo in the question's time value.