Q.
The $x$-$t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = \frac{4}{3}$ s is
A) $\frac{\sqrt{3}}{32}\ \pi^{2}cms^{- 2}$
B) $- \frac{\pi^{2}}{32}\ cms^{- 2}$
C) $\frac{\pi^{2}}{32}\ cms^{- 2}$
D) $- \frac{\sqrt{3}}{32}\ \pi^{2}cms^{- 1}$
Explanation
(d) $T = 8s,\ \omega = \frac{2\pi}{T} = \left( \frac{\pi}{4} \right){rads}^{- 1}$
$x = A\sin{\omega\ t}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = - \omega^{2}x = \ - \left( \frac{\pi^{2}}{16} \right)\sin\left( \frac{\pi}{4}t \right)$
Substituting $t = \frac{4}{3}$ s, we get
$a = \ - \left( \frac{\sqrt{3}}{32}\pi^{2} \right)cms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.