If displacement $x$ and velocity $v$ are related as $4v^2 = 25 -x^2$ in a $SHM$. Then time Period of given $SHM$ is (Consider $SI\, units$)
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$4 v^{2}=25-x^{2} \Rightarrow \frac{x^{2}}{25}+\frac{4 v^{2}}{25}=1$
Comparing with $\frac{\mathrm{x}^{2}}{\mathrm{A}^{2}}+\frac{\mathrm{v}^{2}}{(\mathrm{A} \omega)^{2}}=1$
$A=5, A \omega=\frac{5}{2} \quad \therefore \omega=\frac{1}{2} \Rightarrow T=4 \pi$
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