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

Options

A. `cos^-1((7)/(9))`
B. `cos^-1((5)/(9))`
C. `cos^-1((4)/(9))`
D. `cos^-1((1)/(9))`

Explanation

To solve the problem, we need to find the phase difference between two particles moving parallel to the x-axis with the same amplitude and frequency. Let's break down the solution step by step.

### Step 1: Understand the Situation
We have two particles moving along the x-axis with the same amplitude \( A \) and frequency \( \omega \). At a certain instant, they are at positions \( +\frac{A}{3} \) and \( -\frac{A}{3} \) respectively. Their velocities are in the same direction.

### Step 2: Write the Position Equations
For simple harmonic motion, the position of the particles can be described by:
- Particle 1: \( x_1 = A \sin(\omega t) \)
- Particle 2: \( x_2 = A \sin(\omega t + \phi) \)

Where \( \phi \) is the phase difference we need to find.

### Step 3: Set Up the Equations
From the problem, we know:
- For Particle 1: \( x_1 = \frac{A}{3} \)
- For Particle 2: \( x_2 = -\frac{A}{3} \)

Substituting these into the equations:
1. \( \frac{A}{3} = A \sin(\omega t) \)
2. \( -\frac{A}{3} = A \sin(\omega t + \phi) \)

### Step 4: Simplify the Equations
From the first equation:
\[ \sin(\omega t) = \frac{1}{3} \]

From the second equation:
\[ \sin(\omega t + \phi) = -\frac{1}{3} \]

### Step 5: Use the Sine Addition Formula
Using the sine addition formula:
\[ \sin(\omega t + \phi) = \sin(\omega t) \cos(\phi) + \cos(\omega t) \sin(\phi) \]

Substituting the known values:
\[ -\frac{1}{3} = \left(\frac{1}{3}\right) \cos(\phi) + \cos(\omega t) \sin(\phi) \]

### Step 6: Find \( \cos(\omega t) \)
To find \( \cos(\omega t) \), we can use the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \cos(\omega t) = \sqrt{1 - \sin^2(\omega t)} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

### Step 7: Substitute Back
Now substitute \( \cos(\omega t) \) back into the equation:
\[ -\frac{1}{3} = \left(\frac{1}{3}\right) \cos(\phi) + \left(\frac{2\sqrt{2}}{3}\right) \sin(\phi) \]

### Step 8: Rearranging the Equation
Multiply through by 3 to eliminate the denominators:
\[ -1 = \cos(\phi) + 2\sqrt{2} \sin(\phi) \]

### Step 9: Solve for \( \phi \)
Rearranging gives:
\[ \cos(\phi) + 2\sqrt{2} \sin(\phi) + 1 = 0 \]

This is a linear equation in terms of \( \sin(\phi) \) and \( \cos(\phi) \). We can solve this equation to find \( \phi \).

### Step 10: Determine Possible Values of \( \phi \)
After solving the equation, we find that the possible values for \( \phi \) are:
- \( \phi = \cos^{-1}\left(\frac{7}{9}\right) \) or \( \phi = 180^\circ \)

However, since both particles are moving in the same direction, \( \phi = 180^\circ \) is not a valid solution.

### Final Answer
Thus, the phase difference between the two particles is:
\[ \phi = \cos^{-1}\left(\frac{7}{9}\right) \]

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