A particle stars SHM at time t=0. Its amplitude is A and angular frequency is `omega.` At time t=0 its kinetic energy is `E/4`. Assuming potential energy to be zero and the particle can be written as (E=total mechanical energy of oscillation).

Options

A. x=A sin `omega t +pi//6`
B. x=A sin `(omega t +5pi/6)`
C. x=A sin `(omega t +7pi/6)`
D. x=A sin `(omega t +4pi/6)`

Answer

D

Explanation

`(1)/(2)momega^(2)x^(2)=(E)/(4)=(1)/(4)xx(1)/(2)momega^(2)A^(2)`
`Rightarrow x=_(-)^(+)(A)/(2) Rightarrow v=_(-)^(+)(omegaA)/(2)Rightarrowx=_(-)^(+)(sqrt3A))/(2)`
`phi= (pi)/(3),(2pi)/(3),(4pi)/(3),(5pi)/(3) Rightarrow x=(A)/(2)sin(omegat+phi)` where `phi= (pi)/(3),(2pi)/(3),(4pi)/(3),(5pi)/(3)`