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. `asqrt(7)//4`
C. `sqrt(3)//a2`
D. `3a//4`

Explanation

To solve the problem step by step, we will analyze the motion of the two particles A and B performing Simple Harmonic Motion (SHM) and find the required displacement.

### Step 1: Understanding the Setup
Both particles A and B perform SHM along the same straight line with:
- Amplitude = \( a \)
- Frequency = \( f \)
- Equilibrium position = \( O \)

The greatest distance between them is given as \( \frac{3a}{2} \).

### Step 2: Maximum Displacement Analysis
Since the maximum distance between the two particles is \( \frac{3a}{2} \), we can express this in terms of their displacements from the equilibrium position. Let the displacements of A and B from the mean position O be \( x_A \) and \( x_B \) respectively.

The maximum distance can be expressed as:
\[ |x_A - x_B| = \frac{3a}{2} \]

### Step 3: Equal Frequency and Amplitude
Since both particles have the same amplitude and frequency, we can assume that at the point of maximum separation, they are at the following positions:
- Particle A is at \( x_A = \frac{3a}{4} \)
- Particle B is at \( x_B = -\frac{3a}{4} \)

This gives:
\[ | \frac{3a}{4} - (-\frac{3a}{4}) | = \frac{3a}{2} \]

### Step 4: Finding the Condition for Same Displacement
At some instant, both particles have the same displacement from the mean position. This means:
\[ x_A = x_B \]
Let’s denote this common displacement as \( x \).

### Step 5: Using Trigonometric Relationships
From the triangle formed by the displacements, we can use trigonometric identities. We know:
\[ \sin \theta = \frac{3}{4} \]
From this, we can find \( \cos \theta \):
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16} \]
Thus,
\[ \cos \theta = \frac{\sqrt{7}}{4} \]

### Step 6: Finding the Displacement
The displacement \( x \) can be expressed in terms of the amplitude \( a \) and \( \cos \theta \):
\[ x = a \cos \theta = a \cdot \frac{\sqrt{7}}{4} \]

### Final Answer
Thus, the displacement from the mean position is:
\[ x = \frac{\sqrt{7} a}{4} \]