For a particle executing simple harmonic motion, the displacement-time $(x-t)$ graph is as shown in the figure. The acceleration of the particle at $t=\frac{4}{3} \mathrm{~s}$ is

• Correct answer
  A

  $-\frac{\sqrt{3}}{32} \pi^2 \mathrm{~cm} \mathrm{~s}^{-2}$
• B

  $\frac{32}{\sqrt{3}} \pi^2 \mathrm{~cm} \mathrm{~s}^{-2}$
• C

  $+\frac{\sqrt{3}}{32} \pi \mathrm{cm} \mathrm{s}^{-2}$
• D

  $+\frac{32}{\sqrt{3}} \pi \mathrm{cm} \mathrm{s}^{-2}$

Solution

The displacement-time graph shown in the figure is a sine wave, so the equation of displacement, $x=1 \sin \omega t$
Here, $\quad T=8 \mathrm{~s}$ (for a complete cycle) Hence, $\omega=\frac{2 \pi}{T}=\frac{2 \pi}{8}=\frac{\pi}{4} \mathrm{rad} / \mathrm{s}$ and acceleration in SHM, $$ \begin{aligned} \alpha & =-\omega^2 a \sin \omega t \\ \alpha & =-\left(\frac{\pi}{4}\right)^2 \times 1 \times \sin \left(\frac{\pi}{4} \times \frac{4}{3}\right) \\ \Rightarrow \quad \alpha & =-\frac{\pi^2}{16} \times \frac{\sqrt{3}}{2}=-\frac{\pi^2 \sqrt{3}}{32} \mathrm{~cm} \mathrm{~s}^{-2} \end{aligned} $$ Hence, the correct option is (a).

Hence, the correct answer is option: 'A'.