Shows the displacement time graph of a particle executing SHM with a time period `T`. Four points `1,n 2, 3` and `4` are market on the graph where the displacement is half that of the amplitude.

a. Identify the point of same displacement but with opposite direction of motion . Find the time difference between them.
b. Identify the point where the particles move in the same difference. Find the time difference between them.

Explanation

a.Points `1 and 2` have same displacement `x = + A//2` with equal but opposite velocities.
Similarly `3 and 4` have same displacement `x = - A//2` with equal but opposite velocities
The phase difference between `1 and 2 and 3 and 4` is
`Delta phi = (2)/(3) pi`
Therefore , the time difference is
`Delta t = (Delta f)/(omega) = - (T)/(2 pi) = (T)/(3)`
b.Points `1 and 4` have same direction of motion . Similarly points `2 and 3` have same direction of motion.
The Phase difference between `1 and 4` is
`Delta phi = 2 pi - 2 ((pi)/(6)) = (5 pi)/(3)`
Therefore , the time difference is `Delta t = (Delta phi)/(omega) = (T)/(2 pi) = (5 T)/(3)`
The plase difference between `2 and 3` is `Delta phi = 2 ((pi)/(6)) = (pi)/(3)`
Thus the time difference is `Delta t = (Delta phi)/(omega) = (T)/(2 pi) = ((pi)/(3)) = (T)/(6)`