Question

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 $t=\frac{4}{3}s$ is

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

Answer 1

Answer: (A)

The displacement-time graph shown in the figure is a sine wave,
so the equation of
displacement, $x=1\sin \omega t$

Here, $T=8s$ (for a complete cycle)
Hence, $\omega =\frac{2\pi }{T}=\frac{2\pi }{8}=\frac{\pi }{4}rad/s$
and acceleration in SHM,
$\alpha =-{\omega }^{2}a\sin \omega t$
$\alpha =-{\left(\frac{\pi }{4}\right)}^2\times 1\times \sin \left(\frac{\pi }{4}\times \frac{4}{3}\right)$
$\Rightarrow \alpha =-\frac{{\pi }^2}{16}\times \frac{\sqrt{3}}{2}$
$=-\frac{{\pi }^2\sqrt{3}}{32}cm{s}^{-2}$