The potential energy of a simple harmonic oscillator when the particle is half way to its end point is
(where, E is the total energy)

Options

A. `(U_(max))/(2)`
B. `(U_(max))/(3)`
C. `(U_(max))/(4)`
D. `2 U_(max)`

Answer

C

Explanation

PE at `x = (A)/(2)` is `(1)/(2) m omega^(2) xx (A^(2))/(4)`
`therefore" "PE = (1)/(4)[(1)/(2)m omega^(2)A^(2)]=(U_(max))/(4)`