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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 the right with the vertical, the other pendulum makes an angle of $1^o$ to the left of the vertical. What is the phase difference between the pendulums ?
Solution
Given situation are shown in given below figures $(i)$ and $(ii)$
Suppose both the pendulums follows the below functions,
$\theta_{1}=\theta_{0} \sin \left(\omega t+\phi_{1}\right)$
$\theta_{2}=\theta_{0} \sin \left(\omega t+\phi_{2}\right)$
$\cdots$ $(1)$
$\cdots(2)$
where, $\theta_{0}=$ amplitude
For first pendulum at any time $t, \quad \theta_{1}=+\theta_{0}$ (Right side)
$\therefore$ From equation $(1)$
$+\theta_{0}=\theta_{0} \sin \left(\omega t+\phi_{1}\right)$
$\therefore+1=\sin \left(\omega t+\phi_{1}\right)$
$\therefore \omega t+\phi_{1}=\frac{\pi}{2}$
Similarly for second pendulum at time $t$,
$\theta_{2}=-\frac{\theta_{0}}{2} \quad(\text { left side })$
$\therefore$ From equation $(2)$
$-\frac{\theta_{0}}{2}=\theta_{0} \sin \left(\omega t+\phi_{2}\right)$
$\therefore-\frac{1}{2}=\sin \left(\omega t+\phi_{2}\right)$
$\therefore \omega t+\phi_{2}=-\frac{\pi}{6} \text { or } \frac{7 \pi}{6}$
