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

Answer 1

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

Velocity of particle executing S.H.M $v=\omega \sqrt{a^2-x^2}$ and 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 $x=2cm$

$v=|a|$

$\Rightarrow {\omega }^2\times 2=\omega \sqrt{3^2-2^2}$

$\Rightarrow \omega =\frac{\sqrt{5}}{2}$

$\Rightarrow \frac{2\pi }{T}=\frac{\sqrt{5}}{2}$

$\Rightarrow T=\frac{4\pi }{\sqrt{5}}$