Question

The angular velocity and the amplitude of a simple pendulum is $\omega$ and a respectively. At a displacement $x$ from the mean position if its kinetic energy is $T$ and potential energy is $V$, then the ratio of $T$ to $V$ is

• A. $ \frac{ ( a^2 - x^2 \, \omega^2 ) }{ x^2 \, \omega^2 }$
• B. $ \frac{ x^2 \, \omega^2}{ (a^2 - x^2 \, \omega^2 )}$
• C. $ \frac{ (a^2 - x^2) }{ x^2 } $
• D. $ \frac{ x^2 }{ (a^2 - x^2 )} $

Answer 1

Answer: (C)

P.E, $V=\frac{1}{2} m \omega^{2} x^{2}$
and $KE , T=\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)$
$\therefore \frac{T}{V}=\frac{a^{2}-x^{2}}{x^{2}}$