Question

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

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

Answer 1

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

From the given $x-t$ graph of SHM, displacement $=x=A\sin \omega t$
$\therefore v=A\omega \cos \omega t\text{and}a=-A{\omega }^2\sin \omega t$.
Given that,
$A=1\text{cm},\omega =\frac{2\pi }{8}{\text{rad s}}^{-1}$ and $t=\frac{4}{3}\text{s}$
$\Rightarrow a=-\left(1\text{cm}\right){\left(\frac{2\pi }{8}{s}^{-1}\right)}^2\sin (\frac{2\pi }{8}.\frac{4}{3})=-\frac{\sqrt{3}{\pi }^2}{32}{\text{cm s}}^{-2}$