Two particle `A` and `B` execute `SHM` along the same line with the same amplitude a, same frequency and same equlibrium position `O`. If the phase difference between them is `phi=2 sin^(-1)(0.9)` then find maxium distance between two.

Answer

To find the maximum distance between two particles A and B executing simple harmonic motion (SHM) with a phase difference, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
Both particles A and B have the same amplitude \( A \), frequency, and equilibrium position. Their displacement equations can be written as:
\[ x_1 = A \sin(\omega t) \]
\[ x_2 = A \sin(\omega t + \phi) \]
where \( \phi = 2 \sin^{-1}(0.9) \).

2. **Finding the Displacement Difference**:
The distance between the two particles is given by:
\[ d = x_1 - x_2 = A \sin(\omega t) - A \sin(\omega t + \phi) \]
Factoring out \( A \):
\[ d = A \left( \sin(\omega t) - \sin(\omega t + \phi) \right) \]

3. **Using the Sine Difference Formula**:
We can use the sine difference formula:
\[ \sin(c) - \sin(d) = 2 \cos\left(\frac{c+d}{2}\right) \sin\left(\frac{c-d}{2}\right) \]
Applying this to our equation:
\[ d = A \cdot 2 \cos\left(\frac{2\omega t + \phi}{2}\right) \sin\left(\frac{-\phi}{2}\right) \]

4. **Calculating \(\phi\)**:
We know:
\[ \phi = 2 \sin^{-1}(0.9) \]
Therefore,
\[ \sin\left(\frac{\phi}{2}\right) = 0.9 \]
Thus,
\[ \sin\left(-\frac{\phi}{2}\right) = -0.9 \]

5. **Substituting Values**:
Now substituting back into the distance equation:
\[ d = A \cdot 2 \cos\left(\frac{2\omega t + \phi}{2}\right) \cdot (-0.9) \]
Simplifying gives:
\[ d = -1.8 A \cos\left(\frac{2\omega t + \phi}{2}\right) \]

6. **Finding Maximum Distance**:
The maximum distance occurs when \( \cos\left(\frac{2\omega t + \phi}{2}\right) = \pm 1 \):
\[ d_{\text{max}} = | -1.8 A | = 1.8 A \]

### Final Answer:
The maximum distance between the two particles A and B is:
\[ \boxed{1.8 A} \]