The velocity of a wave propagating along a stretched string is 10 m/s and its frequency is 100 Hz. The phase difference between the particles situated at a distance of 2.5 cm on the string will be-

Options

A. `pi//8`
B. `pi//4`
C. `3pi//8`
D. `pi//2`

Answer

To solve the problem step by step, we will follow the concepts of wave motion, specifically focusing on the relationship between wave velocity, frequency, wavelength, and phase difference.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Velocity of the wave, \( v = 10 \, \text{m/s} \)
- Frequency of the wave, \( f = 100 \, \text{Hz} \)
- Distance between the particles, \( \Delta x = 2.5 \, \text{cm} = 0.025 \, \text{m} \)

2. **Calculate Wavelength (\( \lambda \)):**
The relationship between wave velocity (\( v \)), frequency (\( f \)), and wavelength (\( \lambda \)) is given by:
\[ v = f \lambda \]
Rearranging this formula to find the wavelength:
\[ \lambda = \frac{v}{f} = \frac{10 \, \text{m/s}}{100 \, \text{Hz}} = 0.1 \, \text{m} = 10 \, \text{cm} \]

3. **Calculate the Phase Constant (\( k \)):**
The phase constant \( k \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
Substituting the value of \( \lambda \):
\[ k = \frac{2\pi}{0.1 \, \text{m}} = 20\pi \, \text{rad/m} \]

4. **Calculate the Phase Difference (\( \Delta \phi \)):**
The phase difference between two points separated by a distance \( \Delta x \) is given by:
\[ \Delta \phi = k \Delta x \]
Substituting the values of \( k \) and \( \Delta x \):
\[ \Delta \phi = 20\pi \, \text{rad/m} \times 0.025 \, \text{m} = 0.5\pi \, \text{rad} \]

5. **Final Result:**
The phase difference between the particles situated at a distance of 2.5 cm on the string is:
\[ \Delta \phi = \frac{\pi}{2} \, \text{rad} \]