Question

A particle executes SHM on a straight line path. The amplitude of oscillation is $2cm$. When the displacement of the particle from the mean position is $1cm$, the numerical value of magnitude of acceleration is equal to the numerical value of magnitude of velocity. Then find out the frequency of SHM.

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

Answer 1

Answer: (A)

The speed of particle at a distance $x=1$ from mean position is
$v=\omega \sqrt{A^2-x^2}=\omega \sqrt{2^2-1^2}=\sqrt{3}s$ .......(i)
The magnitude of acceleration at $x=1$ is
$a={\omega }^{2}x={\omega }^2$ .......(ii)
from equation (i) and (ii)
${\omega }^2=\sqrt{3}\omega \Rightarrow \omega =\sqrt{3}$
or $f=\frac{\omega }{2\pi }=\frac{\sqrt{3}}{2\pi }$