The `KE` and `PE` , of a particle executing `SHM` amplitude `A` will be equal when its displacement is

Options

A. `Asqrt(2)`
B. `A//2`
C. `A//sqrt(2)`
D. `Asqrt(2//3)`

Answer

C

Explanation

`K = (1)/(2) m omega ^(2)(A^(2)- y^(2))`
`U = (1)/(2) m omega^(2)y^(2)`
`K = U` or `(1)/(2) m omega^(2) (A^(2) - y^(2)) = (1)/(2) m omega^(2) y^(2)`
`i.e. 2y^(2) = A^(2)` or `y = (A)/(sqrt(2))`