For a particle executing `SHM`, the kinetic energy `(K)` is given by `K = K_(0)cos^(2) omegat`. The equation for its displacement is

Options

A. `((K_(0))/(m omega^(2)))^(1/2) sin omega t`
B. `((2K_(0))/(m omega^(2)))^(1/2) sin omega t`
C. `((2omega^(0))/(m K_(0)))^(1/2) sin omega t`
D. `((2K_(0))/(m omega))^(1/2) sin omega t`

Answer

B

Explanation

`K_(0) = (1)/(2) m V_(max)^(2)`
`K_(0) = (1)/(2)m xx A^(2)omega^(2)`
`A = ((2K_(0))/(m omega^(2)))^(1//2)`
Equation for displacement `y = A sin omega t`
`= ((2K_(0))/(m omega^(2)))^(1//2) sin omega t`