Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of `2^(@)` to the right with the vertical, the other pendulum makes an angle of `1^(@)` to the left of the vertical. What is the phase difference between the pendulums?

Explanation

Given situation are shown in given below figures (i) and (ii)

Suppose both the pendulums follows the below functions,
`theta_(1) = theta_(0) sin (omega t+ phi_(1))" ""…….."(1)`
`theta_(2) = theta_(0) sin (omega t+phi_(2))" ""........."(2)`
where, `theta_(0)=` amplitude
For first pendulum at any time t,
`theta_(1)= +theta_(0)" "` (Right side)
From equation (1)
`+ theta_(1)= theta_(0) sin (omega t+phi_(1))`
`therefore +1 = sin(omega t+phi_(1))`
`therefore omega t + phi_(1) = (pi)/(2)" "".........."(3)`
Similarly for second pendulum at time t,
`theta_(2) = -(theta_0)/(2)" "` (left side)
From equation (2)
`-(theta_0)/(2)= theta_(0) sin (omega t+ phi_(2))`
`therefore -(1)/(2)= sin (omega t+ phi_(2))`
`therefore omega +phi_(2)= -(pi)/(6)" or "(7pi)/(6)" ""......."(4)[i.e. = 2pi-(pi)/(6)]`
the difference in phases from equ. (3) and (4)
`(omega + phi_(2))-(omega t-phi_(1))= (7pi)/(6)-(pi)/(2)`
`therefore phi_(2)-phi_(1)= (4pi)/(6)= (2pi)/(3)`
`therefore phi_(2)-phi_(1)= 120^(@)`.