A particle of `m` is executing `SHM`about the origin on x- axis frequencxy `sqrt((ka)/(pi m))`, where `k` is a constant and a is the amplitude Find its potential energy if `x` is the displecement at time t:

Options

A. `kax^(2)`
B. `ka^(2)x`
C. `2pi kax^(2)`
D. `2pi kx^(2)`

Answer

C

Explanation

`2pi f = omega, 2pi sqrt((ka)/(pi m)) = omega`
Potential energy `U = (1)/(2) m omega^(2)x^(2) = 2pi ka x^(2)`