A progressive wave of frequency 550 Hz is travelling with a velocity of 360 ms How far apart are the two points `60^@` out of phase ?

Answer

To solve the problem of finding the distance between two points that are 60 degrees out of phase in a progressive wave, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency (f) = 550 Hz
- Velocity (v) = 360 m/s
- Phase difference (Δφ) = 60 degrees

2. **Convert Phase Difference to Radians:**
- Since the formula we will use requires the phase difference in radians, we convert degrees to radians:
\[ \Delta \phi \text{ (in radians)} = \Delta \phi \text{ (in degrees)} \times \frac{\pi}{180} \]
\[ \Delta \phi = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \text{ radians} \]

3. **Calculate the Wavelength (λ):**
- The wavelength can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
\[ \lambda = \frac{360 \, \text{m/s}}{550 \, \text{Hz}} = \frac{360}{550} \approx 0.6545 \, \text{m} \]

4. **Use the Phase Difference to Find Distance (Δx):**
- The relationship between phase difference and distance is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Rearranging this to find Δx gives:
\[ \Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} \]
Substituting the values:
\[ \Delta x = \frac{\frac{\pi}{3} \cdot 0.6545}{2\pi} \]
Simplifying:
\[ \Delta x = \frac{0.6545}{6} \approx 0.1091 \, \text{m} \]

5. **Final Answer:**
- The distance between the two points that are 60 degrees out of phase is approximately:
\[ \Delta x \approx 0.109 \, \text{m} \]