Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.

Explanation

If the amplitude of oscillator is A and displacement x at time t, then total energy
`E= (1)/(2) kA^(2)` and
Potential energy `U= (1)/(2) kx^(2)`
but potential energy `PE= (1)/(2) E` is given,
`therefore (1)/(2) kx^(2)= (1)/(2)xx (1)/(2) kA^(2)`
`therefore x^(2)= (A^2)/(2)`
`therefore x= pm (A)/(sqrt(2))`.
Sign `pm` indicates that oscillator will be in one side from mean position.