The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of `1/N` of its amplitude from the mean position is

Options

A. `N^(2)+1`
B. `1/N^2`
C. `N^2`
D. `N^2-1`

Answer

D

Explanation

The kinetic energy of the body executing simple harmonic motion at a distance x from the mean position is
`K=1/2 momega^2 (A^2-x^2)`
and the potential energy is `U=1/2momega^2x^2`
where A is the amplitude , `omega` is the angular frequency and m is the mass of the body
At `x=A/N, K=1/2 momega^2 (A^2-(A/N)^2)`
`U=1/2momega^2 (A/N)^2`
Their corresponding ratio is = `=((A^2-A^2/N^2))/(A^2/N^2)=(A^2/N^2(N^2-1))/(A^2/N^2)`
`=N^2-1`