NEET Physics 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: (1991)

$ \frac{(a^2 - x^2 \omega^2)}{x^2 \omega^2} $

$ \frac{x^2 \omega^2}{(a^2 - x^2 \omega^2)} $

$ \frac{(a^2 - x^2)}{x^2} $

$ \frac{x^2}{(a^2 - x^2)} $

Solution:

Kinetic energy is $ T = \frac{1}{2}m\omega^2(a^2 - x^2) $ and potential energy is $ V = \frac{1}{2}m\omega^2x^2 $. Taking the ratio gives $ \frac{T}{V} = \frac{a^2 - x^2}{x^2} $.

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