Two particle `P` and `Q` describe `S.H.M.` of same amplitude a same frequency `f` along the same straight line .The maximum distance between the two particles is `asqrt(2)` The phase difference between the two particle is

Options

A. zero
B. `(pi)/(2)`
C. `(pi)/(6)`
D. `(pi)/(3)`

Explanation

To find the phase difference between two particles \( P \) and \( Q \) that are performing simple harmonic motion (SHM) with the same amplitude \( A \) and frequency \( f \), we start by analyzing their displacement equations.

### Step-by-Step Solution:

1. **Define the Displacement Equations**:
- For particle \( P \):
\[ y_1 = A \sin(\omega t) \]
- For particle \( Q \):
\[ y_2 = A \sin(\omega t + \phi) \]
Here, \( \phi \) is the phase difference we need to find.

2. **Calculate the Difference in Displacement**:
- The difference in displacement between the two particles is given by:
\[ y_2 - y_1 = A \sin(\omega t + \phi) - A \sin(\omega t) \]
- Factoring out \( A \):
\[ y_2 - y_1 = A \left( \sin(\omega t + \phi) - \sin(\omega t) \right) \]

3. **Use the Sine Difference Identity**:
- We can apply the trigonometric identity:
\[ \sin a - \sin b = 2 \cos\left(\frac{a + b}{2}\right) \sin\left(\frac{a - b}{2}\right) \]
- Let \( a = \omega t + \phi \) and \( b = \omega t \):
\[ y_2 - y_1 = A \cdot 2 \cos\left(\frac{(\omega t + \phi) + \omega t}{2}\right) \sin\left(\frac{(\omega t + \phi) - \omega t}{2}\right) \]
- This simplifies to:
\[ y_2 - y_1 = 2A \cos\left(\omega t + \frac{\phi}{2}\right) \sin\left(\frac{\phi}{2}\right) \]

4. **Set the Maximum Distance**:
- We know the maximum distance between the two particles is given as \( A\sqrt{2} \):
\[ |y_2 - y_1| = A\sqrt{2} \]
- Therefore, we can equate:
\[ 2A \cos\left(\omega t + \frac{\phi}{2}\right) \sin\left(\frac{\phi}{2}\right) = A\sqrt{2} \]

5. **Simplify the Equation**:
- Dividing both sides by \( A \):
\[ 2 \cos\left(\omega t + \frac{\phi}{2}\right) \sin\left(\frac{\phi}{2}\right) = \sqrt{2} \]
- Rearranging gives:
\[ \cos\left(\omega t + \frac{\phi}{2}\right) \sin\left(\frac{\phi}{2}\right) = \frac{\sqrt{2}}{2} \]

6. **Maximize the Cosine Term**:
- The maximum value of \( \cos\left(\omega t + \frac{\phi}{2}\right) \) is \( 1 \). For this to hold, we need:
\[ \sin\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]
- This implies:
\[ \frac{\phi}{2} = \frac{\pi}{4} \quad \Rightarrow \quad \phi = \frac{\pi}{2} \]

### Conclusion:
The phase difference \( \phi \) between the two particles \( P \) and \( Q \) is:
\[ \phi = \frac{\pi}{2} \]