Question

A particle executes linear simple harmonic motion with amplitude of $2cm$. When the particle is at $1cm$ from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is ________________.

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

Answer 1

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

let x= 2sin($\omega t$)

then $v_x=2\omega cos(\omega t$)

and $a_x=-2{\omega }^{2}sin\left(\omega t\right)$

If x=1 then sin($\omega t$) = 0.5

or ($\omega t$) = $\frac{\pi }{6}$

put this value in eqn 2 and eqn 3 and then equate both

we get $\omega t= -cot($\omega t$)

or $\omega =\sqrt{3}{s}^{-1}$

T = $\frac{2\pi }{\sqrt{3}}$ sec