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?

Consider a pair of identical pendulums, which oscillate with equal amp

1.	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?
Answer» Solution :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^(@)`.