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. `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

`(KE)/(PE)=((1)/(2)momega^(2)(A^(2)-x^(2)))/((1)/(2)momega^(2)x^(2))`
`implies (T)/(V)=(A^(2)-X^(2))/(X^(2))=(a^(2)-X^(2))/(X^(2))`.