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

Options

A. `Asqrt2`
B. `(A)/(2)`
C. `(A)/(sqrt2)`
D. `Asqrt((2)/(3))`

Answer

C

Explanation

`K=(1)/(2)momega^2(A^2-y^2)`
`U=(1)/(2)momega^2y^2`
`K+U=(1)/(2)momega^2(A^2-y^2)+(1)/(2)momega^2y^2`
i.e., `2y^2=A^2` or `y=(A)/(sqrt2)`