The maximum seperation between two particles executing simple harmonic motion of the equal amplitude and same frequency along the same straight line and about the same mean position, is `sqrt(3)` times the amplitude of oscillation. IF the phase difference between them is `phi`, determine `2pi//phi`.

Explanation

To solve the problem, we need to determine the phase difference \( \phi \) between two particles executing simple harmonic motion (SHM) given that their maximum separation is \( \sqrt{3} \) times the amplitude \( A \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Let the displacements of the two particles be represented as:
\[ x_1 = A \sin(\omega t) \]
\[ x_2 = A \sin(\omega t + \phi) \]
- Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase difference between the two particles.

2. **Finding the Separation**:
- The separation \( x \) between the two particles is given by:
\[ x = x_2 - x_1 = A \sin(\omega t + \phi) - A \sin(\omega t) \]
- Using the sine subtraction formula, we can express this as:
\[ x = A \left( \sin(\omega t + \phi) - \sin(\omega t) \right) \]

3. **Using the Sine Difference Formula**:
- The difference of sines can be expressed as:
\[ \sin(A) - \sin(B) = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
- Applying this to our equation:
\[ x = A \cdot 2 \cos\left(\frac{2\omega t + \phi}{2}\right) \sin\left(\frac{\phi}{2}\right) \]
- Thus, the maximum separation occurs when \( \cos\left(\frac{2\omega t + \phi}{2}\right) = 1 \):
\[ x_{\text{max}} = 2A \sin\left(\frac{\phi}{2}\right) \]

4. **Setting Up the Equation**:
- According to the problem, the maximum separation is given as:
\[ x_{\text{max}} = \sqrt{3} A \]
- Therefore, we equate the two expressions:
\[ 2A \sin\left(\frac{\phi}{2}\right) = \sqrt{3} A \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ 2 \sin\left(\frac{\phi}{2}\right) = \sqrt{3} \]

5. **Solving for \( \sin\left(\frac{\phi}{2}\right) \)**:
- Rearranging gives:
\[ \sin\left(\frac{\phi}{2}\right) = \frac{\sqrt{3}}{2} \]
- The angle whose sine is \( \frac{\sqrt{3}}{2} \) is:
\[ \frac{\phi}{2} = \frac{\pi}{3} \quad \text{(or } 60^\circ\text{)} \]

6. **Finding \( \phi \)**:
- Multiplying both sides by 2:
\[ \phi = \frac{2\pi}{3} \]

7. **Calculating \( \frac{2\pi}{\phi} \)**:
- Finally, we compute:
\[ \frac{2\pi}{\phi} = \frac{2\pi}{\frac{2\pi}{3}} = 3 \]

### Final Answer:
\[ \frac{2\pi}{\phi} = 3 \]