A particle executes `SHM` with an amplitude of `10cm` and frequency `2 Hz`. At `t = 0`, the particle is at a point where potential energy and kinetic energy are same. The equation for its displacement is

Options

A. `x = 0.1 sin (4 pi t + (pi)/(4))m`
B. `x = 0.1 (sin 4 pi )m`
C. `x = 0,1 cos (4pi t + (pi)/(3))m`
D. `x = 0.1 (sin 4 pi t - (pi)/(3))m`

Answer

A

Explanation

`U = K = (1)/(2)E`
`= (1)/(2) ((1)/(2) m (A omega)^(2))`
`v^(2) = (1)/(2) A^(2) omega^(2)`
`omega^(2) (A^(2) - x^(2)) = (1)/(2) A^(2) omega^(2) rArr x = (A)/(sqrt(2))` at t = 0
`x = A sin (omega t + phi)`
`(A)/(sqrt(2)) = A sin phi = phi = (pi)/(4)`