The maximum potential energy of a simple harmonic oscilator is `U_("max")`. The the P.E. of the oscillator when it is half way to its end point is

Options

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

Answer

C

Explanation

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