If the displacement (x) and velocity (v) of a particle executing simple harmonic motion are related through the expression $4v^2=25-x^2$, then its time period is

Given $4v^2=25-x^2$. Differentiating with respect to time t, we have
$8v\frac{dv}{dt}=0-2x\frac{dx}{dt}$
or $8va=-2xv \left(\because a=\frac{dv}{dt},v=\frac{dx}{dt}\right)$
where a is the acceleration. Thus
$a=-\left(\frac{1}{4}\right)x$        (i)
For a simple harmonic motion
$a=-{\omega }^{2}x$          (ii)
Comparing (i) and (ii), we have
$\omega =\frac{1}{2} \text{ or } \frac{2\pi }{T}=\frac{1}{2}$
or $T=4\pi$