Question

A particle executes linear simple harmonic motion with an amplitude of 2$cm$ . When the particle is at 1$cm$ 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. $\frac{2\pi }{3}$
• D. $\frac{\sqrt{3}}{2\pi }$

Answer 1

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

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

and acceleration $={\omega }^{2}x$

Now given,

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

$\Rightarrow {\omega }^2.1=\omega \sqrt{2^2-1^1}$

$\Rightarrow \omega =\sqrt{3}$

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