figure show the identical simple pan-delums of length. One is tilled at an angle `alpha` and imparted an initial velocity `v_(1)` towards mean possition and at a velocity `v_(2)` at an initial angular displacement `beta`. Find the phase difference in oscillations of these two pendulums

Explanation

It is given that first pendulum bob is given a velocity `v_(1)` at a dislpacement `l alpha` from mean position , using the formula for velocity we can find the amplitude of its oscillation as
`V_(1) = omega sqrt(A_(1)^(2) - (l alpha)^(2))` [If `A_(1)`, is the amplitude of SHM of this As for simple pendulum `omega = sqrt((g)/(l))` we have
`V_(1)^(2) = (g)/(l) [(A_(1)^(2)) - (l alpha)^(2))]`
`A_(1) = sqrt(l^(2) alpha^(2) + v_(1)^(2)l)/(g)` (i)
Similarly if `A_(2)` is the amplitude of SHM of second pendlum bob we have
`V_(2) = omega sqrt(A_(1)^(2) - (l beta)^(2))`
or `A_(2) = sqrt((l^(2) beta^(2) + (v_(2)^(2)l)/(g))` (ii)
Now we repersent the two SHMs by circular motion representation as shown in figure

In first pendulum at `t = 0` the bob is thrown from a displacement `l alpha` from mean position with a velocity `v_(1)` towards mean position . As it is moving towards mean position in figure (a) , we consider the corresponding circular motion particle `A_(0)` of the bob A is second quarant, as the referance direction of `omega` we consider anticlockwise. As shown in figure initial phase of bob A is given as
`phi_(1) = pi - sin^(-1) ((l alpha)/(A_(1)))` (iii)
Similarly for second pendulum bob B, its corresponding circular motion partical `B_(0) at t = 0` is consider as shown in figure (b) its initial base is given as
`phi_(2) = sin^(-1) ((l beta)/(A_(2)))` (iv)
As both the pandulum are identical, their angular frequency for SHM must be same , so their phase difference can be given as
`Delta phi = phi_(2) - phi_(1) sin^(-1) ((l beta)/(A_(2))) + sin^(-1) ((l alpha)/(A_(1))) - pi`