Question

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 1

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

Assuming the two pendulums follow the following functions of their angular displacements

`θ_1 = θ_0 sin(ωt + phi_1)`  .......(i)

And `θ_2 = θ_0 sin(ωt + phi_2)`  .......(ii)

As it is given amplitude and time period are equal but phases being different.

Now, for the first pendulum at any time t 

`θ_1 = + θ_0`   .....[Right extreme]

From equation (i), we get

⇒ `θ_0 = θ_0 sin(ωt + phi_1)` or 1 = `sin(θt + phi_1)` 

⇒ `sin  pi/2 = sin(ωt + phi_1)`

or `(ωt + phi_1) = pi/2`  ......(iii)

Similarly, at the same instant t for pendulum second, we have `θ_2 = - θ_0/2` where θ₀ = 2° is the angular amplitude of the first pendulum. For the second pendulum, the angular displacement is one degree, therefore `θ_2 = θ_0/2` and a negative sign is taken to show for being left to mean position.

From equation (ii), then `- θ_0/2 = θ_0 sin(ωt + phi_2)`

⇒ `sin(ωt + phi_2) = - 1/2`

⇒ `(ωt + phi_2) = - pi/6 or (7pi)/6`

or `(ωt + phi_2) = - pi/6 or (7pi)/6`  ......(iv)

From equations (iv) and (iii), the difference in phases

`(ωt + phi_2) - (ωt + phi_2) = (7pi)/6 - pi/2`

= `(7pi - 3pi)/6`

= `(4pi)/6`

Or `(phi_2 - phi_1) = (4pi)/6 = (2pi)/3` = 120°