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