The amgular 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 U, then the ratio of T to U is

Options

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^(2))`
D. `(x^(2))/((a^(2)-x^(2)))`

Answer

C

Explanation

`P.E.,V=(1)/(2)m omega^(2)x^(2)" and K.E., T"=(1)/(2)m omega^(2)(a^(2)-x^(2))`
`:." "(T)/(V)=(a^(2)-x^(2))/(x^(2))`