Question

A particle is executing S.H.M. and its velocity $v$ is related to its position $(x)$ as $v^{2}+a x^{2}=b$, where $a$ and $b$ are positive constants. The frequency of oscillation of particle is

• A. $\frac{1}{2 \pi} \sqrt{\frac{b}{a}}$
• B. $\frac{\sqrt{a}}{2 \pi}$
• C. $\frac{\sqrt{b}}{2 \pi}$
• D. $\frac{1}{2 \pi} \sqrt{\frac{a}{b}}$

Answer 1

Answer: (B)

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