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

$4{\mathrm{v}}^2=25-{\mathrm{x}}^2,$
Differentiating both sides with respect to t
$$\begin{gathered} 4\left(2\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dt}}\right)=-2\mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}} \\ \Rightarrow 4\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{x}\Rightarrow 4\mathrm{a}=-\mathrm{x} \\ \Rightarrow \mathrm{a}=-\frac{1}{4}\mathrm{x}\Rightarrow {\mathrm{\omega }}^2=\frac{1}{4} \\ \Rightarrow \mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=4\mathrm{\pi } \end{gathered}$$

$$\begin{gathered} \mathrm{other} \mathrm{method} : \\ \mathrm{comparing} \mathrm{v}=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2} \\ \mathrm{here}, \mathrm{v}=\frac{1}{2}\sqrt{25-{\mathrm{x}}^{2 }}\Rightarrow \mathrm{\omega }=\frac{1}{2} \\ \Rightarrow \mathrm{T}=4\mathrm{\pi } \end{gathered}$$