If displacement x and velocity y are related as $4{\mathrm{v}}^2 = 25-{\mathrm{x}}^2$ in a SHM Then time period of given SHM is (consider SI units)

Given $4{\mathrm{v}}^2 = 5-{\mathrm{x}}^2$

Differentiating with respect to t on both sides,

$8\mathrm{va} = -2\mathrm{xv} \Rightarrow \mathrm{a} = -\frac{\mathrm{x}}{4}$

Motion is simple harmonic:

So, ${\omega }^2 =\frac{1}{4} and T = 2\mathrm{\pi }\sqrt{4} = 4\mathrm{\pi }$