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 frequency of SHM (in second-1) is:

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

Answer 1

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

It is given that

$\left|v\right|=\left|a\right|\Rightarrow \omega \sqrt{A^2-x^2}={\omega }^{2}x\Rightarrow \omega =\frac{\sqrt{A^2-x^2}}{x}\Rightarrow 2\pi f=\frac{\sqrt{A^2-x^2}}{x}\Rightarrow f=\frac{1}{2\pi }\left(\frac{\sqrt{A^2-x^2}}{x}\right)=\frac{1}{2\pi }\left(\frac{\sqrt{2^2-1^2}}{1}\right)=\frac{\sqrt{3}}{2\pi }$