Question

A particle executes 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. Then its time period in seconds is

• A. $\frac{\sqrt{5}}{2\pi }$
• 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 of particle executing S.H.M
$v=\omega \sqrt{a^2-x^2}$
Acceleration, $a=-{\omega }^{2}x$
$x$ = displacement at any time interval
$a$ = amplitude$=3cm$
$\omega$ = angular frequency $=\frac{2\pi }{T}$
$T$ = Time period
Now, at $x=2cm$
$v=|a|$
$\Rightarrow {\omega }^2\times 2=\omega \sqrt{3^2-2^2}$
$\Rightarrow \omega =\frac{\sqrt{5}}{2}$
$\Rightarrow \frac{2\pi }{T}=\frac{\sqrt{5}}{2}$
$\Rightarrow T=\frac{4\pi }{\sqrt{5}}s$