Two particles execute SHM of same amplitude and same time period, about same mean position but with a phase difference between them. At an instant they are found to cross each other at `x=+(A)/(3)`. The phase difference between them is

Options

A. `2cos^(-1)((1)/(5))`
B. `2sin^(-1)((1)/(5))`
C. `2cos^(-1)((1)/(3))`
D. `2sin^(-1)((1)/(5))`

Explanation

To solve the problem, we need to find the phase difference between two particles executing Simple Harmonic Motion (SHM) with the same amplitude and time period, crossing each other at a position \( x = \frac{A}{3} \).

### Step-by-Step Solution:

1. **Understanding the SHM Equations**:
The position of a particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Position of Both Particles**:
Let particle 1 have a phase \( \phi_1 \) and particle 2 have a phase \( \phi_2 \). Their positions at any time \( t \) can be expressed as:
\[ x_1 = A \sin(\omega t + \phi_1) \]
\[ x_2 = A \sin(\omega t + \phi_2) \]

3. **Condition for Crossing**:
The particles cross each other at \( x = \frac{A}{3} \). Therefore, we have:
\[ A \sin(\omega t + \phi_1) = \frac{A}{3} \]
\[ A \sin(\omega t + \phi_2) = \frac{A}{3} \]

4. **Simplifying the Equations**:
Dividing both equations by \( A \):
\[ \sin(\omega t + \phi_1) = \frac{1}{3} \]
\[ \sin(\omega t + \phi_2) = \frac{1}{3} \]

5. **Finding the Angles**:
From the above equations, we can express the angles:
\[ \omega t + \phi_1 = \sin^{-1}\left(\frac{1}{3}\right) \quad \text{(1)} \]
\[ \omega t + \phi_2 = \sin^{-1}\left(\frac{1}{3}\right) \quad \text{(2)} \]

6. **Phase Difference Calculation**:
The phase difference \( \Delta \phi \) between the two particles is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
From equations (1) and (2):
\[ \Delta \phi = \left(\omega t + \phi_2\right) - \left(\omega t + \phi_1\right) = \phi_2 - \phi_1 \]

7. **Using the Sine Function**:
Since both particles are at the same position \( \frac{A}{3} \), we can also express \( \phi_2 \) in terms of \( \phi_1 \):
\[ \phi_2 = \pi - \phi_1 \quad \text{(since } \sin(\pi - x) = \sin(x)\text{)} \]

8. **Final Phase Difference**:
Therefore, substituting \( \phi_2 \):
\[ \Delta \phi = \left(\pi - \sin^{-1}\left(\frac{1}{3}\right)\right) - \sin^{-1}\left(\frac{1}{3}\right) = \pi - 2\sin^{-1}\left(\frac{1}{3}\right) \]

9. **Using Inverse Trigonometric Identity**:
We can use the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \):
\[ \Delta \phi = 2\cos^{-1}\left(\frac{1}{3}\right) \]

### Conclusion:
The phase difference \( \Delta \phi \) between the two particles is:
\[ \Delta \phi = 2\cos^{-1}\left(\frac{1}{3}\right) \]