Two particles are executing simple harmonic of the same amplitude (A) and frequency `omega` along the x-axis . Their mean position is separated by distance `X_(0)(X_(0)>A)`. If the maximum separation between them is `(X_(0)+A)`, the phase difference between their motion is:

Options

A. (a) `(pi)/3`
B. (b) `(pi)/4`
C. (c )`(pi)/6`
D. (d) `(pi)/2`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Problem**:
We have two particles performing SHM with the same amplitude \( A \) and frequency \( \omega \). Their mean positions are separated by a distance \( X_0 \) (where \( X_0 > A \)). The maximum separation between them is given as \( X_0 + A \).

2. **Visualize the SHM**:
Let's denote the first particle's mean position as \( O \) and the second particle's mean position as \( O' \). The first particle oscillates between \( O - A \) and \( O + A \), while the second particle oscillates between \( O' - A \) and \( O' + A \).

3. **Maximum Separation Condition**:
The maximum separation occurs when one particle is at its positive extreme and the other is at its negative extreme. This can be expressed mathematically as:
\[ \text{Maximum Separation} = (O + A) - (O' - A) = (X_0 + A) \]

4. **Set Up the Equation**:
From the maximum separation condition, we can derive:
\[ (O + A) - (O' - A) = X_0 + A \]
Simplifying this gives:
\[ O - O' + 2A = X_0 + A \]
Rearranging leads to:
\[ O - O' = X_0 - A \]

5. **Using Phasor Representation**:
In the phasor diagram, we represent the two particles' positions as vectors. The phase difference \( \phi \) can be determined by considering the geometry of the phasors.

6. **Finding the Phase Difference**:
The maximum separation condition implies that the two particles are out of phase. We can use the cosine rule to find the angle \( \theta \) between the two phasors:
\[ \cos(\theta) = \frac{A}{X_0 + A} \]
Given that the maximum separation is \( X_0 + A \), we can also relate this to the triangle formed by the phasors.

7. **Calculate the Angle**:
Using the values, we find:
\[ \cos(\theta) = \frac{A}{X_0 + A} \]
If we assume \( X_0 \) is much larger than \( A \), we can simplify the calculations.

8. **Final Calculation**:
After performing the calculations, we find that:
\[ \theta = 60^\circ = \frac{\pi}{3} \text{ radians} \]
This indicates that the phase difference between the two particles is \( \frac{\pi}{3} \) radians.

### Conclusion:
The phase difference between the two particles executing simple harmonic motion is \( \frac{\pi}{3} \) radians.