Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. Their phase difference is

Options

A. `(5pi)/6`
B. `(4pi)/3`
C. `pi/6`
D. `(2pi)/3`

Answer

To find the phase difference between two particles executing simple harmonic motion (SHM) and passing each other at half their amplitude, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two particles, P and Q, executing SHM with the same amplitude (A) and frequency along parallel lines.
- They pass each other moving in opposite directions when their displacement is half their amplitude (A/2).

2. **Displacement of Particles**:
- For particle P, when it is at displacement A/2, it is moving towards the mean position from the maximum displacement (A).
- For particle Q, when it is at displacement A/2, it is moving towards the maximum displacement from the mean position.

3. **Finding the Angle for Particle P**:
- The displacement of particle P can be expressed as:
\[ y_P = A \sin(\theta_P) \]
- Setting \(y_P = A/2\):
\[ A \sin(\theta_P) = \frac{A}{2} \]
- Simplifying gives:
\[ \sin(\theta_P) = \frac{1}{2} \]
- Therefore, \(\theta_P = 30^\circ\) (or \(\pi/6\) radians).

4. **Finding the Angle for Particle Q**:
- The displacement of particle Q can similarly be expressed as:
\[ y_Q = A \sin(\theta_Q) \]
- Since particle Q is moving towards the maximum position, we can express its position as:
\[ y_Q = A \sin(\theta_Q) = -\frac{A}{2} \]
- This gives:
\[ \sin(\theta_Q) = -\frac{1}{2} \]
- Therefore, \(\theta_Q = -30^\circ\) (or \(-\pi/6\) radians).

5. **Calculating the Phase Difference**:
- The phase difference \(\phi\) between the two particles is given by:
\[ \phi = \theta_Q - \theta_P \]
- Substituting the values:
\[ \phi = -30^\circ - 30^\circ = -60^\circ \]
- To express this in a positive form, we can add \(360^\circ\):
\[ \phi = 360^\circ - 60^\circ = 300^\circ \]
- Alternatively, we can express it as:
\[ \phi = 180^\circ - 60^\circ = 120^\circ \]

6. **Final Result**:
- The phase difference between the two particles is:
\[ \phi = 120^\circ \]