Question

A particle executes SHM with an amplitude of $2cm$. When the particle is at $1cm$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

• A. $\frac{1}{2\pi \sqrt{3}}$
• B. $2\pi \sqrt{3}$
• C. $\frac{2\pi }{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2\pi }$

Answer 1

Answer: (C)

$\frac{2\pi }{\sqrt{3}}$

Given,

$A=2cm$

$x=1cm$

The relation of velocity and displacement is given by

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

Acceleration, $a=-{\omega }^{2}x$

According to question,

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

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

$\frac{2\pi }{T}\times 1=\sqrt{2^2-1^2}=\sqrt{4-1}$

$T=\frac{2\pi }{\sqrt{3}}$

The correct option is C.