Question

If the displacement $(x)$ and velocity $(v)$ of a particle executing SHM are related through the expression $3v^2=30-x^2$, then its time period is _____$s$. (Take $\sqrt{3}=1.73$ )

Answer 1

Answer: 10.86

Given equation,
$3v^2=30-x^2$
Differentiating both sides, we get
$6v\frac{dv}{dx}=-2x$
$\Rightarrow v\frac{dv}{dx}=\frac{-1}{3}x$
$\Rightarrow \frac{a}{x}=\frac{-1}{3}$
$\dots \left[\because a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}\right]$
$T=\frac{2\pi }{\sqrt{\mid \frac{\text{ acceleration }}{\text{ displacement }}}}=\frac{2\pi }{|\frac{-1}{3}|}=2\pi \sqrt{3}$
$\therefore T=10.86sec$