A progressive wave moves with a velocity of 36m/ s in a medium with a frequency of 200Hz. The phase difference between two particles seperated by a distance of 1 cm is

Options

A. `40^@`
B. `20rad `
C. `pi/9 rad `
D. `pi/9` degree

Answer

To solve the problem, we need to find the phase difference between two particles separated by a distance of 1 cm in a progressive wave moving with a velocity of 36 m/s and a frequency of 200 Hz.

### Step-by-Step Solution:

1. **Identify the given values:**
- Velocity of the wave, \( v = 36 \, \text{m/s} \)
- Frequency of the wave, \( f = 200 \, \text{Hz} \)
- Distance between the two particles, \( \Delta x = 1 \, \text{cm} = 0.01 \, \text{m} \)

2. **Calculate the wavelength (\( \lambda \)):**
The relationship between velocity, frequency, and wavelength is given by the formula:
\[ v = f \cdot \lambda \]
Rearranging this gives:
\[ \lambda = \frac{v}{f} \]
Substituting the known values:
\[ \lambda = \frac{36 \, \text{m/s}}{200 \, \text{Hz}} = \frac{36}{200} = 0.18 \, \text{m} \]

3. **Convert the distance to meters:**
The distance \( \Delta x \) is already converted to meters as \( 0.01 \, \text{m} \).

4. **Calculate the phase difference (\( \phi \)):**
The phase difference between two points separated by a distance \( \Delta x \) is given by:
\[ \phi = \frac{2\pi \Delta x}{\lambda} \]
Substituting the values we have:
\[ \phi = \frac{2\pi \cdot 0.01 \, \text{m}}{0.18 \, \text{m}} \]
Simplifying this:
\[ \phi = \frac{2\pi \cdot 0.01}{0.18} = \frac{2\pi}{18} = \frac{\pi}{9} \, \text{radians} \]

5. **Final Result:**
The phase difference between the two particles separated by a distance of 1 cm is:
\[ \phi = \frac{\pi}{9} \, \text{radians} \]