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. The time period of SHM (in Hz) is:

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

Answer 1

The correct option is B $\frac{2\pi }{\sqrt{3}}$

${\omega }^2\left(1\right)=\omega \sqrt{(4-1)}\Rightarrow \omega =\sqrt{3}\Rightarrow T=\frac{2\pi }{\sqrt{3}}s$