Two particle execute `SHM` of same amplitude of `20 cm` with same period along the same line about the same equilibrium position. The maximum distance between the two is `20 cm`. Their phase difference in radians is

Options

A. `2pi//3`
B. `pi//2`
C. `pi//3`
D. `pi//4`

Answer

To find the phase difference between two particles executing simple harmonic motion (SHM) with the given conditions, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have two particles executing SHM with the same amplitude (A = 20 cm) and the same period. The maximum distance between the two particles is given as 20 cm.

2. **Set Up the Equations of Motion**:
- Let the position of the first particle be given by:
\[ X_1 = A \sin(\omega t) \]
- Let the position of the second particle be given by:
\[ X_2 = A \sin(\omega t + \phi) \]
where \(\phi\) is the phase difference we need to find.

3. **Calculate the Distance Between the Two Particles**:
- The distance \(D\) between the two particles at any instant is:
\[ D = |X_2 - X_1| = |A \sin(\omega t + \phi) - A \sin(\omega t)| \]
- Using the trigonometric identity for the difference of sines, we can express this as:
\[ D = A \left| \sin(\omega t + \phi) - \sin(\omega t) \right| = 2A \sin\left(\frac{\phi}{2}\right) \cos\left(\omega t + \frac{\phi}{2}\right) \]

4. **Find the Maximum Distance**:
- The maximum distance occurs when \(\cos\left(\omega t + \frac{\phi}{2}\right) = 1\):
\[ D_{\text{max}} = 2A \sin\left(\frac{\phi}{2}\right) \]
- We know from the problem that \(D_{\text{max}} = 20 \, \text{cm}\) and \(A = 20 \, \text{cm}\):
\[ 20 = 2 \times 20 \sin\left(\frac{\phi}{2}\right) \]

5. **Solve for \(\sin\left(\frac{\phi}{2}\right)\)**:
- Simplifying the equation:
\[ 20 = 40 \sin\left(\frac{\phi}{2}\right) \]
\[ \sin\left(\frac{\phi}{2}\right) = \frac{20}{40} = \frac{1}{2} \]

6. **Find \(\frac{\phi}{2}\)**:
- The angle for which \(\sin\left(\frac{\phi}{2}\right) = \frac{1}{2}\) is:
\[ \frac{\phi}{2} = \frac{\pi}{6} \quad \text{(or 30 degrees)} \]

7. **Calculate \(\phi\)**:
- Therefore:
\[ \phi = 2 \times \frac{\pi}{6} = \frac{\pi}{3} \quad \text{(or 60 degrees)} \]

### Final Answer:
The phase difference \(\phi\) between the two particles is:
\[ \phi = \frac{\pi}{3} \text{ radians} \]