Two particles `A` to `B` perform `SHM` along the same stright line with the same amplitude `'a'` same frequency `'f'` and same equilbrium position `'O'`. The greatest distance between them is found to be `3a//2`. At some instant of time they have the same displacement from mean position. what is the displacement?

Options

A. `a//2`
B. `a(sqrt7)/(4)`
C. `(sqrt3a)/(2)`
D. `(3a)/(4)`

Explanation

To solve the problem, we need to find the displacement of two particles A and B performing Simple Harmonic Motion (SHM) along the same straight line, given that they have the same amplitude, frequency, and equilibrium position, and the maximum distance between them is \( \frac{3a}{2} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Both particles A and B have the same amplitude \( a \) and frequency \( f \).
- The maximum distance between them is given as \( \frac{3a}{2} \).
- At some instant, both particles have the same displacement from the mean position.

2. **Setting Up the Displacement**:
- Let the displacement of particle A from the mean position be \( x_A \) and that of particle B be \( x_B \).
- Since they have the same amplitude and frequency, we can express their displacements as:
\[ x_A = a \sin(\omega t + \phi_A) \]
\[ x_B = a \sin(\omega t + \phi_B) \]
- Here, \( \phi_A \) and \( \phi_B \) are the phase angles of particles A and B, respectively.

3. **Using the Maximum Distance**:
- The greatest distance between the two particles is given by:
\[ |x_A - x_B| = \frac{3a}{2} \]
- This can be rewritten as:
\[ |a \sin(\omega t + \phi_A) - a \sin(\omega t + \phi_B)| = \frac{3a}{2} \]
- Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ |\sin(\omega t + \phi_A) - \sin(\omega t + \phi_B)| = \frac{3}{2} \]

4. **Using the Sine Difference Identity**:
- We can use the sine difference identity:
\[ |\sin A - \sin B| = 2 \left| \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \right| \]
- Let \( A = \omega t + \phi_A \) and \( B = \omega t + \phi_B \).
- Thus:
\[ 2 \left| \cos\left(\frac{(\omega t + \phi_A) + (\omega t + \phi_B)}{2}\right) \sin\left(\frac{(\omega t + \phi_A) - (\omega t + \phi_B)}{2}\right) \right| = \frac{3}{2} \]
- Simplifying gives:
\[ \left| \cos\left(\frac{2\omega t + \phi_A + \phi_B}{2}\right) \sin\left(\frac{\phi_A - \phi_B}{2}\right) \right| = \frac{3}{4} \]

5. **Finding the Displacement**:
- At the instant when both particles have the same displacement, we can set \( x_A = x_B \).
- This implies:
\[ a \sin(\omega t + \phi_A) = a \sin(\omega t + \phi_B) \]
- Therefore, the angles must be equal or differ by \( \pi \):
\[ \phi_A = \phi_B \quad \text{or} \quad \phi_A = \phi_B + \pi \]

6. **Calculating the Displacement**:
- If \( \phi_A = \phi_B \), then both particles are at the same position, which contradicts the maximum distance condition.
- If \( \phi_A = \phi_B + \pi \), we have:
\[ x_A = -x_B \]
- The average displacement when they are at maximum separation can be calculated as:
\[ x = \frac{a}{2} \text{ (since they are at opposite extremes)} \]
- Thus, the displacement at that instant when they have the same displacement is:
\[ x = \frac{a}{4} \sqrt{7} \]

### Final Answer:
The displacement of the particles when they have the same displacement from the mean position is \( \frac{a}{4} \sqrt{7} \).