The angular velocity and the amplitude of a simple pendulum is w 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

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:

1. \((a^2 - x^2)^2\)
2. \((a^2 - x^2) \omega^2\)
3. \((a^2 - x^2) / x^2\)
4. \(2 / x^2\)

Detailed Answer:

For a simple pendulum, the total energy is: \[ E = \frac{1}{2} m \omega^2 a^2 \] The potential energy \( V \) at displacement \( x \) is: \[ V = \frac{1}{2} m \omega^2 x^2 \] The kinetic energy \( T \) is the difference between total energy and potential energy: \[ T = E - V = \frac{1}{2} m \omega^2 (a^2 - x^2) \] The ratio of kinetic energy to potential energy is: \[ \frac{T}{V} = \frac{\frac{1}{2} m \omega^2 (a^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{a^2 - x^2}{x^2} \] Correct answer: (c) \((a^2 - x^2) / x^2\).