Question

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


1) 12

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

Answer 1

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

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 $t=\frac{4}{3}\text{s}$
$\Rightarrow a=-\left(1\text{cm}\right){\left(\frac{2\pi }{8}{s}^{-1}\right)}^2\sin (\frac{2\pi }{8}.\frac{4}{3})=-\frac{\sqrt{3}{\pi }^2}{32}{\text{cm s}}^{-2}$