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

Question

Aicharya · Accepted Answer

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)

For a particle executing simple harmonic motion, the displacement-time $(x-t)$ graph is as shown in the…

Solution