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^(@)` 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?

Answer

120

Explanation

Assuming the two pendulums follow the following functions of their displacement :
`theta_(1) = theta_(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)
From Eq. (i), we get:
`rArr theta_(0) = theta_(0) sin(omegat + phi_(1))`
or `1 = sin (omegat + phi_(1))`
or `(omegat + phi_(1)) = pi/2`............. (iii)
Similarly, at the same instant t for pendulum second, we have
`theta_(2) = -theta_(0)/2`
Note `theta_(0) = 2^(@)` is the angular amplitude for the second pendulum the angular displacement is one degree therefore, `theta_(2) = theta_(0)/2` and negative sign is taken to show for being left to mean position.
From Eq. (ii), then
`- theta_(2)/2 = theta_(0)sin(omegat + phi_(2))`
`rArr sin(omegat + phi_(2)) = -1/2`
`rArr (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` or `pi/2 = (7pi- 3pi)/6 = (4pi)/6`
or `(phi_(2) - phi_(1)) = (4pi)/6 = (2pi)/3 = 120^(@)`