The angular velocity and amplitude of simple pendulum are `omega` and r respectively. At a displacement x from the mean position, if its kinetic energy is T and potential energy is U, find the ration of T to U.

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

For a simple pendulum, oscillating with very small angular displacement, the oscillations are treated as linear SHM's. We can apply the formulae of linear SHM.
For a linear SHM, `x= a "sin" omega t`
For a displacemetn x, its P.E. and K.E. are given by
`U=(1)/(2) m omega^(2)x^(2)`
and `T=(1)/(2) m omega^(2)(a^(2)-x^(2))`
`therefore (T)/(U)=(a^(2)-x^(2))/(x^(2))`