Question

A particle exhibits linear simple harmonic motion with an amplitude of $3cm$. When the particle is at $2cm$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Its time period in seconds is

• A. $\sqrt{\frac{5}{2\pi }}$
• B. $\frac{4\pi }{\sqrt{5}}$
• C. $\frac{2\pi }{\sqrt{3}}$
• D. $\frac{\sqrt{5}}{\pi }$

Answer 1

The correct option is B $\frac{4\pi }{\sqrt{5}}$

Velocity and acceleration in an SHM are given by

$v=\omega \sqrt{A^2-x^2}$
$a=-{\omega }^{2}x$

Given, $v=\omega \sqrt{9-4}$
$a=-{\omega }^2\times 2$

and $a=v$
$\Rightarrow \omega \sqrt{9-4}=-{\omega }^2\times 2$
$\Rightarrow \omega =\frac{\sqrt{5}}{2}$
$\Rightarrow T=\frac{2\pi }{\omega }=\frac{4\pi }{\sqrt{5}}$