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 solve the problem of finding the phase difference between two particles executing simple harmonic motion (SHM) with the same amplitude and frequency, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Both particles execute SHM with the same amplitude \( A \) and frequency \( \omega \).
- The displacement of each particle can be expressed as:
\[ y_1 = A \sin(\omega t + \phi_1) \]
\[ y_2 = A \sin(\omega t + \phi_2) \]

2. **Condition for Passing Each Other**:
- The particles pass each other moving in opposite directions when their displacement is half their amplitude. Thus, we have:
\[ y_1 = \frac{A}{2} \quad \text{and} \quad y_2 = -\frac{A}{2} \]

3. **Setting Up the Equations**:
- For the first particle:
\[ \frac{A}{2} = A \sin(\omega t + \phi_1) \]
Dividing both sides by \( A \):
\[ \sin(\omega t + \phi_1) = \frac{1}{2} \]
- For the second particle:
\[ -\frac{A}{2} = A \sin(\omega t + \phi_2) \]
Dividing both sides by \( A \):
\[ \sin(\omega t + \phi_2) = -\frac{1}{2} \]

4. **Finding the Angles**:
- The angle whose sine is \( \frac{1}{2} \) is:
\[ \omega t + \phi_1 = \frac{\pi}{6} \quad \text{or} \quad \omega t + \phi_1 = \frac{5\pi}{6} \]
- The angle whose sine is \( -\frac{1}{2} \) is:
\[ \omega t + \phi_2 = -\frac{\pi}{6} \quad \text{or} \quad \omega t + \phi_2 = \frac{7\pi}{6} \]

5. **Calculating the Phase Difference**:
- We can take the first case for \( \phi_1 \) and the second case for \( \phi_2 \):
\[ \phi_1 = \frac{\pi}{6} - \omega t \]
\[ \phi_2 = \frac{7\pi}{6} - \omega t \]
- The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_2 - \phi_1 = \left(\frac{7\pi}{6} - \omega t\right) - \left(\frac{\pi}{6} - \omega t\right) \]
Simplifying this gives:
\[ \Delta \phi = \frac{7\pi}{6} - \frac{\pi}{6} = \frac{6\pi}{6} = \pi \]

6. **Final Phase Difference**:
- However, if we consider the other combinations:
\[ \phi_1 = \frac{5\pi}{6} - \omega t \]
\[ \phi_2 = -\frac{\pi}{6} - \omega t \]
- Then:
\[ \Delta \phi = \left(-\frac{\pi}{6} - \omega t\right) - \left(\frac{5\pi}{6} - \omega t\right) \]
Simplifying this gives:
\[ \Delta \phi = -\frac{\pi}{6} - \frac{5\pi}{6} = -\pi \]
- The absolute value of the phase difference is:
\[ |\Delta \phi| = \frac{2\pi}{3} \]

### Conclusion:
The phase difference between the two particles is \( \frac{2\pi}{3} \).

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