Q. 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) ${(a}^{2} - x^{2}\omega^{2})/x^{2}\omega^{2}$
B) $x^{2}\omega^{2}{/(a}^{2} - x^{2}\omega^{2})$
C) ${(a}^{2} - x^{2})/x$
D) ${x^{2}/(a}^{2} - x^{2})$
Explanation
(c) PE, $V = \frac{1}{2}m\omega^{2}x^{2}$
and KE, $T = \frac{1}{2}m\omega^{2}(a^{2} - x^{2})$
$\therefore\frac{T}{V} = \ \frac{a^{2} - x^{2}}{x^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.