Question

The x-t graph of a particle undergoing simple harmonic motion is as shown in the figure.
The acceleration of the particle at $t=\frac{4}{3}s$ is the

• A. $\frac{\sqrt{3}}{32}{\pi }^{2}cm{s}^{-2}$
• B. $-\frac{{\pi }^2}{32}cm{s}^{-2}$
• C. $\frac{{\pi }^2}{32}cm{s}^{-2}$
• D. $-\frac{\sqrt{3}}{32}{\pi }^{2}cm{s}^{-2}$.

Answer 1

The correct option is D $-\frac{\sqrt{3}}{32}{\pi }^{2}cm{s}^{-2}$.

From graph,
$A=1cm,T=8s$
$x=Asin\omega t=Asin\frac{2\pi }{T}t$
At $t=\frac{4}{3}s$
$x=1sin\frac{2\pi }{8}\times \frac{4}{3}=sin\frac{\pi }{3}=\frac{\sqrt{3}}{2}cm$

In SHM,
acceleration $=-{\omega }^{2}x=-\frac{4{\pi }^2}{T^2}\times \frac{\sqrt{3}}{2}cm{s}^{-2}=-\frac{4{\pi }^2}{(8{)}^2}\times \frac{\sqrt{3}}{2}=-\frac{\sqrt{3{\pi }^2}}{32}cm{s}^{-2}$