The equation of sound wave travelling along negative X-direction is given by, ` y= 0.04 sin pi (500t +1.5x)m.`The shortest distance between two particles having phase difference of `pi ` at the same instant is

Options

A. 0.66 m
B. 0.5 m
C. 0.33 m
D. 0.2 m

Answer

To find the shortest distance between two particles having a phase difference of \( \pi \) at the same instant in the given wave equation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Wave Equation**:
The wave equation provided is:
\[ y = 0.04 \sin(\pi (500t + 1.5x)) \]
This can be rewritten as:
\[ y = 0.04 \sin(500\pi t + 1.5\pi x) \]

2. **Determine the Wave Parameters**:
From the equation, we can identify:
- Angular frequency \( \omega = 500\pi \) rad/s
- Wave number \( k = 1.5\pi \) rad/m

3. **Calculate the Wavelength \( \lambda \)**:
The relationship between wave number \( k \) and wavelength \( \lambda \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
Substituting \( k = 1.5\pi \):
\[ 1.5\pi = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{1.5\pi} = \frac{2}{1.5} = \frac{4}{3} \text{ m} \approx 0.66 \text{ m} \]

4. **Determine the Shortest Distance for Phase Difference \( \pi \)**:
The phase difference \( \Delta \phi \) between two points in a wave is related to the distance \( d \) by the equation:
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot d \]
For a phase difference of \( \pi \):
\[ \pi = \frac{2\pi}{\lambda} \cdot d \]
Rearranging gives:
\[ d = \frac{\pi \cdot \lambda}{2\pi} = \frac{\lambda}{2} \]
Substituting \( \lambda = 0.66 \text{ m} \):
\[ d = \frac{0.66}{2} = 0.33 \text{ m} \]

5. **Conclusion**:
The shortest distance between two particles having a phase difference of \( \pi \) at the same instant is:
\[ \boxed{0.33 \text{ m}} \]