Question

The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t\, = \,4/3\, s$ is

• A. $\frac{\sqrt3}{32}\pi^2cm/s^{2}$
• B. $\frac{-\pi^2}{32}cm/s^{2}$
• C. $\frac{\pi^2}{32}cm/s^{2}$
• D. $-\frac{\sqrt3}{32}\pi^2 cm/s^{2}$

Answer 1

Answer: (D)

$T=8 s, \omega=\frac{2 \pi}{T}=\left(\frac{\pi}{4}\right) rads ^{-1}$
$\Rightarrow x=A \sin \omega t$
$\therefore a=-\omega^{2} x=-\left(\frac{\pi^{2}}{16}\right) \sin \left(\frac{\pi}{4} t\right)$
Substituting $t=\frac{4}{3} s$, we get
$a=-\left(\frac{\sqrt{3}}{32} \pi^{2}\right) cms ^{-2}$