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

Explanation

Consider the situations shown in the diagram (i) and (ii)

Assuming the two pendulums follows the following functions of their angular displacements
`theta_(1)=0_(0)sin(omegat+phi_(1))" " .....(i)`
and `theta_(2)=theta_(0)sin(omegat+phi_(2))" " .....(ii)`
As it is given that amplitude and time period being equal but phases being different. Now, for first pendulum at any time t
`theta_(1)=+theta_(0)` [Right extreme]
Form Eq. (i), we get
`rArrtheta_(0)=theta_(0)sin(omegat+phi_(1))" ro "1=sin(omegat+phi_(1))`
`rArrsin""(pi)/(2)=sin(omegat+phi_(1))`
or `(omegat+phi_(1))=(pi)/(2)`
Similary, at the same instant t for pendulum second, we have
`theta_(2)=-(theta_(0))/(2)`
where `theta_(2)=2^(@)` is the angular amplitude of first pendulum. For the second pendulum, the angular displacement is one degree. therefore `theta_(2)=(theta_(0))/(2)` and negative sign i s taken to show
for being left to mean position.
From Eqs. (ii), then `(theta_(0))/(2)=theta_(0)sin(omegat+phi_(2))`
`rArrsin(omegat+phi_(2))=-(1)/(2)rArr(omegat+phi_(2))=-(pi)/(6)or(7pi)/(6)`
or `(omegat+phi_(2))=-(pi)/(6)or(7pi)/(6)" ".....(iv)`
From Eqs. (iv) and (iii), the difference in phases
`(omegat+phi_(2))-(omegat+phi_(1))=(7pi)/(6)-(pi)/(2)=(7pi-3pi)/(6)=(4pi)/(6)`
or `(phi_(2)-phi_(2))=(4pi)/(6)=(2pi)/(3)=120^(@)`