Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The phase difference is:-

Options

A. `pi/6`
B. 0
C. `(2pi)/(3)`
D. `pi`

Explanation

To find the phase difference between the two oscillating particles, we can follow these steps:

### Step 1: Understand the Problem
We have two particles oscillating with the same frequency and amplitude along parallel lines. They pass each other when their displacement is half of the amplitude (A/2) and are moving in opposite directions.

### Step 2: Write the Equation of Motion
The displacement of a particle in simple harmonic motion (SHM) can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 3: Set Up the Displacement Condition
When the particles pass each other, their displacement is given as:
\[ x = \frac{A}{2} \]
We can set up the equations for both particles. Let's denote the first particle as Particle 1 (red) and the second particle as Particle 2 (blue).

### Step 4: Find the Phase of Particle 1
Assuming Particle 1 starts from the origin, we can express its displacement as:
\[ \frac{A}{2} = A \sin(\omega t_1 + \phi_1) \]
Dividing both sides by \( A \):
\[ \frac{1}{2} = \sin(\omega t_1 + \phi_1) \]
The sine function equals \( \frac{1}{2} \) at:
\[ \omega t_1 + \phi_1 = \frac{\pi}{6} \quad \text{or} \quad \omega t_1 + \phi_1 = \frac{5\pi}{6} \]

### Step 5: Find the Phase of Particle 2
For Particle 2, which is moving in the opposite direction and has completed its motion, we can express its displacement similarly:
\[ \frac{A}{2} = A \sin(\omega t_2 + \phi_2) \]
Again dividing both sides by \( A \):
\[ \frac{1}{2} = \sin(\omega t_2 + \phi_2) \]
For Particle 2, the sine function also equals \( \frac{1}{2} \) at the same angles:
\[ \omega t_2 + \phi_2 = \frac{\pi}{6} \quad \text{or} \quad \omega t_2 + \phi_2 = \frac{5\pi}{6} \]

### Step 6: Determine the Phases
Assuming Particle 1 has a phase of:
\[ \phi_1 = \frac{\pi}{6} \]
Then for Particle 2, which is moving in the opposite direction and has completed its motion, we can take:
\[ \phi_2 = \frac{5\pi}{6} \]

### Step 7: Calculate the Phase Difference
The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the values:
\[ \Delta \phi = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]

### Final Answer
The phase difference between the two particles is:
\[ \Delta \phi = \frac{2\pi}{3} \]

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