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?

Options

A. `(pi)/(2)`
B. `(2)/(3)pi`
C. `(3)/(2)pi`
D. `pi`

Explanation

To find the phase difference between the two pendulums, we will follow these steps:

### Step 1: Define the equations of motion for both pendulums
The angular displacement of a simple harmonic oscillator can be described by the equation:
\[ \theta(t) = \theta_0 \sin(\omega t + \phi) \]
where:
- \(\theta_0\) is the maximum angular displacement (amplitude),
- \(\omega\) is the angular frequency,
- \(t\) is the time,
- \(\phi\) is the phase constant.

### Step 2: Set up the equations for the first pendulum
For the first pendulum, which is at an angle of \(2^\circ\) to the right of the vertical:
\[ \theta_1 = 2^\circ = \theta_0 \sin(\omega t + \phi_1) \]
Given that \(\theta_0 = 2^\circ\):
\[ 2 = 2 \sin(\omega t + \phi_1) \]
Dividing both sides by 2:
\[ 1 = \sin(\omega t + \phi_1) \]
This implies:
\[ \omega t + \phi_1 = \frac{\pi}{2} \quad \text{(Equation 1)} \]

### Step 3: Set up the equations for the second pendulum
For the second pendulum, which is at an angle of \(1^\circ\) to the left of the vertical:
\[ \theta_2 = -1^\circ = \theta_0 \sin(\omega t + \phi_2) \]
Again, using \(\theta_0 = 2^\circ\):
\[ -1 = 2 \sin(\omega t + \phi_2) \]
Dividing both sides by 2:
\[ -\frac{1}{2} = \sin(\omega t + \phi_2) \]
This implies:
\[ \omega t + \phi_2 = \frac{7\pi}{6} \quad \text{(Equation 2)} \]

### Step 4: Find the phase difference
Now, we can find the phase difference \(\phi_2 - \phi_1\) by subtracting Equation 1 from Equation 2:
\[ (\omega t + \phi_2) - (\omega t + \phi_1) = \frac{7\pi}{6} - \frac{\pi}{2} \]
This simplifies to:
\[ \phi_2 - \phi_1 = \frac{7\pi}{6} - \frac{3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]

### Conclusion
The phase difference between the two pendulums is:
\[ \phi_2 - \phi_1 = \frac{2\pi}{3} \]