The angular velocity and the amplitude of a simple pendulum are `omega` and a, respectively. The ratio of its kinetic and potential energies at a displacement x from the mean position is……………

Options

A. `(x^(2)omega^(2))/(a^(2)-x^(2)omega^(2))`
B. `(x^(2))/(a^(2)-x^(2))`
C. `(a^(2)-x^(2)omega^(2))/(x^(2)omega^(2))`
D. `(a^(2)-x^(2))/(x^(2))`

Answer

D

Explanation

Kinetic energy `K=(1)/(2)momega^(2)(a^(2)-x^(2))`
Potential energy `U=(1)/(2)momega^(2)x^(2)implies(K)/(U)=(a^(2)-x^(2))/(x^(2))`