question_answer 59)

                  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 \[{{2}^{o}}\] to die right with die vertical, the other pendulum makes an angle of \[{{1}^{o}}\] to the left of the vertical. What is the phase difference between the pendulums?

Answer:

                  Angular displacement of a pendulum at any instant is given by                 \[\phi ={{\phi }_{0}}\,\,\sin \,(\omega t+\,\theta )\]                 where, \[{{\phi }_{0}}\] is maximum angular displacement and \[\theta \] is the initial phase.                 \[\therefore \] For first pendulum, \[{{2}^{o}}={{2}^{o}}\,\sin \,(\omega t+{{\theta }_{1}})\]                 or \[\sin \,(\omega t+{{\theta }_{1}})=1\] or \[\omega t+{{\theta }_{1}}={{90}^{o}}\]?.. (i)                 For second pendulum, \[-{{1}^{o}}={{2}^{o}}\sin (\omega t+{{\theta }_{2}})\]                 or \[\sin \,(\omega t+{{\theta }_{2}})=-\frac{1}{2}\] or \[(\omega t+{{\theta }_{2}})\]                 \[=-{{30}^{o}}\]                                                ?. (ii)                 From eqns. (i) and (ii),                 \[\omega t+\,{{\theta }_{1}}\,-\omega t\,-{{\theta }_{1}}\,={{120}^{o}}\]                 or \[({{\theta }_{1}}+\,{{\theta }_{2}})\,={{120}^{o}}\]