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 ? 

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 $(2)$

$-\frac{\theta_{0}}{2}=\theta_{0} \sin \left(\omega t+\phi_{2}\right)$

$\therefore-\frac{1}{2}=\sin \left(\omega t+\phi_{2}\right)$

$\therefore \omega t+\phi_{2}=-\frac{\pi}{6} \text { or } \frac{7 \pi}{6}$