The phase difference between two points is `pi//3`. If the frequency of wave is 50 Hz, then what is the distance between two points?
`("Given, "v=330 ms^(-1))`

Options

A. 2.2 m
B. 1.1 m
C. 0.6 m
D. 1.7 m

Answer

To solve the problem, we need to find the distance between two points given the phase difference, frequency, and velocity of the wave.

### Step-by-step Solution:

1. **Understand the relationship between phase difference and path difference:**
The relationship between phase difference (Δφ) and path difference (Δx) is given by the formula:
\[ \Delta x = \frac{\lambda}{2\pi} \Delta \phi \]
where λ is the wavelength and Δφ is the phase difference.

2. **Calculate the wavelength (λ):**
The wavelength can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
where \( v \) is the velocity of the wave and \( f \) is the frequency.

Given:
- \( v = 330 \, \text{m/s} \)
- \( f = 50 \, \text{Hz} \)

Substituting the values:
\[ \lambda = \frac{330 \, \text{m/s}}{50 \, \text{Hz}} = 6.6 \, \text{m} \]

3. **Substitute λ and Δφ into the path difference formula:**
Now, we substitute the calculated wavelength and the given phase difference into the path difference formula. The phase difference is given as:
\[ \Delta \phi = \frac{\pi}{3} \]

Substituting the values:
\[ \Delta x = \frac{6.6 \, \text{m}}{2\pi} \cdot \frac{\pi}{3} \]

4. **Simplify the expression:**
The expression simplifies as follows:
\[ \Delta x = \frac{6.6 \, \text{m}}{2} \cdot \frac{1}{3} = \frac{6.6 \, \text{m}}{6} = 1.1 \, \text{m} \]

5. **Final answer:**
Thus, the distance between the two points is:
\[ \Delta x = 1.1 \, \text{m} \]