Question

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

Answer 1

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