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 os `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?

Explanation

Let `phi_(1)` and `phi_(2)` be the angular displacement of first and second pendulums respectively at an instant. `phi_(0)` be the maximum angular displacement of each pendulum. Let `theta_(1)` and `theta_(2)` be the initial phases of two pendulums, then
`phi_(1)=phi_(0)sin(omegat-theta_(1))` ...(i)
and `phi_(2)=phi_(0)sin(omegat+theta_(2))` ...(ii)
For first pendulum , `phi_(1)=2^(@)=phi_(0)` , From (i), we have
`:. 2=2sin(omegat+theta_(1)) or sin (omegat+theta_(1))=1 or omegat +theta_(1)=90^(@)` ...(iii)
For second pendulum, `phi_(2)=-1^(@),` From (ii), we have
`-1=2sin(omegat +theta_(2))or sin(omegat+theta_(2))=-(1)/(2)=sin(180+30^(@))=sin210^(@)`
`:. omegat+theta_(2)210^(@)` ...(iv)
Subtracting (iii) from (iv), we have
`(omegat+theta_(2))-(omegat+theta_(1))=210-90=120^(@) or theta_(2)-theta_(1)=120^(@)`