A particle starts from point ` x = (-sqrt(3))/(2) `A and move towards negative extreme as shown
(a) Find the equation of the SHM.
(b) Find the time taken by the particle to go directly from its initial position to negative extreme.
(c) Find the time taken by the particle to reach at mean position.

Explanation

Figure shows the solution of the problem with the help of phasor Horizontal component of velocity at Q gives the required direction of velocity at t = 0.
In `Delta` OSQ cos `theta = (sqrt(3)//2A)/(A)=(sqrt(3))/(2) implies theta=(pi)/(6)`
Now `phi = (3n)/(2) -(pi)/(6)=(8pi)/(6)= (4pi)/(3)`
So , equation of SHM is
` x= A sin (omega t + (4 pi)/(3))`
(b) Now to reach the particle at left extreme point it will travel angle `theta` along the circle. So time taken.
`t = (theta )/(omega)= (pi)/(6omega) implies t = (T)/(12) `sec
(c) To reach the particle at mean position it will travel an angle `alpha = (pi)/(2) +(pi)/(6) = (2pi)/(3)`
So, time taken `= (alpha)/(omega) = (T)/(3) `sec