Question

A particle executes SHM on a straight line path. The amplitude of oscillation is $2 \,cm$. When the displacement of the particle from the mean position is $1 \,cm$, 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 $f=\frac{\omega}{2 \pi}=\frac{\sqrt{3}}{2 \pi}$