Two points on a travelling wave having frequency 500 Hz and velocity `300 ms^(-1)` are `60^(@)` out of phase, then the minimum distance between the two point is

Options

A. (a)0.2
B. (b)0.1
C. (c)0.5
D. (d)0.4

Answer

To solve the problem, we need to find the minimum distance between two points on a traveling wave that are 60 degrees out of phase. We are given the frequency and velocity of the wave.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency (f) = 500 Hz
- Velocity (v) = 300 m/s
- Phase difference (Δφ) = 60 degrees

2. **Convert Phase Difference to Radians:**
- Since calculations in physics are often easier in radians, we convert degrees to radians:
\[ \Delta \phi = 60^\circ = \frac{60 \times \pi}{180} = \frac{\pi}{3} \text{ radians} \]

3. **Calculate Wavelength (λ):**
- The wavelength can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
- Substituting the values:
\[ \lambda = \frac{300 \, \text{m/s}}{500 \, \text{Hz}} = \frac{300}{500} = 0.6 \, \text{m} \]

4. **Relate Phase Difference to Path Difference (Δx):**
- The relationship between phase difference and path difference is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
- Rearranging for Δx gives:
\[ \Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} \]

5. **Substituting Values:**
- Now substitute Δφ and λ into the equation:
\[ \Delta x = \frac{\left(\frac{\pi}{3}\right) \cdot (0.6)}{2\pi} \]
- Simplifying this:
\[ \Delta x = \frac{0.6}{6} = 0.1 \, \text{m} \]

6. **Conclusion:**
- The minimum distance between the two points that are 60 degrees out of phase is:
\[ \Delta x = 0.1 \, \text{m} \]

### Final Answer:
The minimum distance between the two points is **0.1 meters**.