$v^2+a x^2=b$
$v^2=b-a x^2$
$v^2=a\left(\frac{b}{a}-x^2\right)$
Comparing it to equation
$v^2=\omega^2\left(A^2-x^2\right)$
$\omega=\sqrt{a}$
$f=\frac{\omega}{2 \pi}=\frac{\sqrt{a}}{2 \pi}$

$y_{2}=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)$
Ratio of amplitude of ${y}_{1}$ to ${y}_{2}={x}: 1$. The value of ${x}$ is ...... .
$y = A\,\cos \,\omega t\,\cos \,2\omega t + A\,\sin \,\omega t\,\sin \,2\omega t$.
Than the nature of the function is
Where $k,k_0,k_1$ and $a$ are all positive