Two particles are in `SHM` with same angular frequency and amplitudes `A` and `2A` respectively along same straight line with same mean position. They cross each other at postion `A//2` distance from mean position in opposite direction. The phase between them is-

Options

A. `(5pi)/(6)-sin^(-1)((1)/(4))`
B. `(pi)/(6)-sin^(-1)((1)/(4))`
C. `(5pi)/(6)-cos^(-1)((1)/(4))`
D. `(pi)/(6)-cos^(-1)((1)/(4))`

Explanation

To find the phase difference between the two particles in simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Motion of the Particles**:
- Let the first particle be described by the equation:
\[ x_1 = A \sin(\omega t) \]
- Let the second particle be described by the equation:
\[ x_2 = 2A \sin(\omega t + \phi) \]
where \( A \) is the amplitude of the first particle, \( 2A \) is the amplitude of the second particle, \( \omega \) is the angular frequency, and \( \phi \) is the phase difference we need to find.

2. **Condition for Crossing**:
- The particles cross each other at a position \( \frac{A}{2} \) from the mean position in opposite directions. Thus, we have:
\[ x_1 = \frac{A}{2} \quad \text{and} \quad x_2 = -\frac{A}{2} \]

3. **Set Up the Equations**:
- For the first particle:
\[ \frac{A}{2} = A \sin(\omega t) \]
- Simplifying this gives:
\[ \sin(\omega t) = \frac{1}{2} \]
- This implies:
\[ \omega t = \frac{\pi}{6} \quad \text{(since } \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\text{)} \]

4. **Substituting Time into the Second Particle's Equation**:
- Now, substitute \( \omega t = \frac{\pi}{6} \) into the second particle's equation:
\[ -\frac{A}{2} = 2A \sin\left(\frac{\pi}{6} + \phi\right) \]
- Dividing both sides by \( 2A \):
\[ -\frac{1}{4} = \sin\left(\frac{\pi}{6} + \phi\right) \]

5. **Finding the Phase Difference**:
- We know that:
\[ \sin\left(\frac{\pi}{6} + \phi\right) = -\frac{1}{4} \]
- To find \( \phi \), we can use the inverse sine function:
\[ \frac{\pi}{6} + \phi = \sin^{-1}\left(-\frac{1}{4}\right) \]
- Rearranging gives:
\[ \phi = \sin^{-1}\left(-\frac{1}{4}\right) - \frac{\pi}{6} \]

6. **Considering the Direction**:
- Since the particles are moving in opposite directions, we can neglect the negative sign:
\[ \phi = -\frac{\pi}{6} - \sin^{-1}\left(\frac{1}{4}\right) \]

7. **Final Result**:
- The phase difference \( \phi \) can be expressed as:
\[ \phi = -\frac{\pi}{6} - \sin^{-1}\left(\frac{1}{4}\right) \]