Physics - SHM Question with Solution | TestHub

In linear simple harmonic motion, the velocity of particle is given by
$v=\omega \sqrt{A^2-x^2}$… (i)
where,$\omega$= angular frequency
A = maximum displacement of amplitude
and x= displacement from mean position.
The acceleration of a particle in simple harmonic motion, (SHM) is given by
$a={\omega }^{2}x$…. (ii)
Here,$x=\frac{A}{2}$
Also,$v=a$(given)
$\omega \sqrt{\left(A^2-x^2\right)}={\omega }^{2}x$[from Eqs. (i) and (ii), we get]
$\Rightarrow \sqrt{\left(A^2-\frac{A^2}{4}\right)}=\omega \times \frac{A}{2}\Rightarrow \frac{\sqrt{3}A}{2}=\omega \times \frac{A}{2}$
$\Rightarrow \frac{2\pi }{T}=\sqrt{3}\mathrm{}\left[\because \omega =\frac{2\pi }{T}\right]$
$\Rightarrow T=\frac{2\pi }{\sqrt{3}}s$