A partical excutes SHM with same frequency and amplitude along the same straight line. They cross each other , at a point midway between the mean and the exterme position. Find the Phase difference between them.

Explanation

Method 1
Let a be the amplitude and `phi` and `phi` there respective phases at any instant ,when they cross each other .
Then their displacement equation can be written as
`x = a sin (phase)`
For the two particals we have `(a)/(2) = a sin phi_(1) and (a)/(2) = a sin phi_(2)`

Taking the least value, `phi_(1) = (pi)/(6)`
and taking the next possible value `phi_(1) = (5 pi)/(6)`
Moreover, from `bar v = a omega cos (phase)`
If `bar v_(1) and bar v_(2)` are the respective velocities of the two particals . The particals cross each other their velocities should be in opposite direction . If we take velocity of the partical `1` positive then velocity of particals should be negative. The velocity of the partical `2` will be negative when it has phase `phi_(1) = 5 pi//6`
Therefore the required phase difference `= phi_(2) - phi_(1) implies (2 pi)/(3) rad`
Method 2
figur show that two particals `P and Q` in SHM along with their corresponding particals in circular motion. Let `P` move in upward direction when crossing `Q at A//2` shown in figure

At htis instant phase of `P` is
`phi_(1) = sin ^(-1) ((1)/(2)) = (pi)/(6)` (i)
Similarly as shown in figure partical `Q` moves in down wards direction (oppositeP) at `A//2`, this implies its circular motion partical is in second quadrant thus its phase angle is
`phi_(2) = pi`
`phi_(1) = pi - sin ^(-1) ((1)/(2)) = pi - (pi)/(6) = (5 pi)/(6)` (ii)
As both are oscillating at same angular frequency their phase difference remain constant which can be given from Eqs (i) and (ii), as `Delta phi = phi_(2) - phi_(1) = (5 pi)/(6) - (pi)/(6) = (2 pi)/(3)`