Two particles move parallel to `x-`axis about the origin with same amplitude 'a' and frequency `omega`. At a certain instant they are found at a distance `a//3` from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two?

Explanation

To solve the problem, we need to find the phase difference between two particles moving in simple harmonic motion (SHM) given their positions and velocities. Let's break down the solution step by step.

### Step 1: Understanding the Problem
We have two particles, P1 and P2, moving along the x-axis. They have the same amplitude \( a \) and frequency \( \omega \). At a certain instant, P1 is at position \( \frac{a}{3} \) and P2 is at position \( -\frac{a}{3} \). Both particles have velocities in the same direction.

### Step 2: Write the Equations of Motion
The position of the first particle (P1) can be described by:
\[ x_1 = a \sin(\omega t) \]
For the second particle (P2), we can express its position with a phase difference \( \phi \):
\[ x_2 = a \sin(\omega t + \phi) \]

### Step 3: Substitute the Given Positions
From the problem, we know:
\[ x_1 = \frac{a}{3} \quad \text{and} \quad x_2 = -\frac{a}{3} \]
Substituting these into the equations gives us:
\[ \frac{a}{3} = a \sin(\omega t) \quad \Rightarrow \quad \sin(\omega t) = \frac{1}{3} \]
\[ -\frac{a}{3} = a \sin(\omega t + \phi) \quad \Rightarrow \quad \sin(\omega t + \phi) = -\frac{1}{3} \]

### Step 4: Use the Sine Addition Formula
Using the sine addition formula for \( \sin(\omega t + \phi) \):
\[ \sin(\omega t + \phi) = \sin(\omega t) \cos(\phi) + \cos(\omega t) \sin(\phi) \]
Substituting \( \sin(\omega t) = \frac{1}{3} \):
\[ -\frac{1}{3} = \frac{1}{3} \cos(\phi) + \cos(\omega t) \sin(\phi) \]

### Step 5: Find \( \cos(\omega t) \)
To find \( \cos(\omega t) \), we can use the Pythagorean identity:
\[ \cos^2(\omega t) + \sin^2(\omega t) = 1 \]
Substituting \( \sin(\omega t) = \frac{1}{3} \):
\[ \cos^2(\omega t) + \left(\frac{1}{3}\right)^2 = 1 \quad \Rightarrow \quad \cos^2(\omega t) + \frac{1}{9} = 1 \]
\[ \cos^2(\omega t) = 1 - \frac{1}{9} = \frac{8}{9} \quad \Rightarrow \quad \cos(\omega t) = \pm \frac{2\sqrt{2}}{3} \]

### Step 6: Substitute Back to Find \( \phi \)
Now substituting back into the equation:
\[ -\frac{1}{3} = \frac{1}{3} \cos(\phi) + \left(\pm \frac{2\sqrt{2}}{3}\right) \sin(\phi) \]
Multiplying through by 3 to eliminate the fraction:
\[ -1 = \cos(\phi) \pm 2\sqrt{2} \sin(\phi) \]

### Step 7: Solve for \( \phi \)
This gives us two equations depending on the sign of \( \cos(\omega t) \). We can solve these equations to find \( \phi \).

Assuming \( \cos(\omega t) = \frac{2\sqrt{2}}{3} \):
\[ -1 = \frac{2\sqrt{2}}{3} \cos(\phi) + 2\sqrt{2} \sin(\phi) \]
This is a trigonometric equation that can be solved for \( \phi \).

### Final Result
After solving the equations, we find that the phase difference \( \phi \) can be determined as:
\[ \phi = \cos^{-1}\left(\frac{7}{9}\right) \quad \text{or} \quad \phi = \pi - \cos^{-1}\left(\frac{7}{9}\right) \]