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. $\left(a^{2}-x^{2} \omega^{2}\right) / x^{2} \omega^{2}$
• B. $x^{2} \omega^{2} /\left(a^{2}-x^{2} \omega^{2}\right)$
• C. $\left(a^{2}-x^{2}\right) / x$
• D. $x^{2} /\left(a^{2}-x^{2}\right)$

Answer 1

Answer: (C)

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