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To calculate the phase difference between two points on a progressive sound wave, we need to follow these steps:

### Given:
- Frequency \( f = 256 \, \text{Hz} \)
- Distance between the two points \( d = 0.16 \, \text{m} \)
- Speed of sound \( v = 330 \, \text{m/s} \)

### Step 1: Calculate the wavelength (\( \lambda \))

The formula for the wavelength is:
\[
\lambda = \frac{v}{f}
\]
Substituting the values:
\[
\lambda = \frac{330 \, \text{m/s}}{256 \, \text{Hz}} = 1.289 \, \text{m}
\]

### Step 2: Calculate the phase difference (\( \Delta \phi \))

The phase difference is given by:
\[
\Delta \phi = \frac{2\pi d}{\lambda}
\]
Substituting the values:
\[
\Delta \phi = \frac{2\pi \times 0.16 \, \text{m}}{1.289 \, \text{m}}
\]

Let me calculate this.The phase difference between the two points 0.16 m apart on the sound wave is approximately \( 0.78 \, \text{radians} \).

Do you need further details or have any questions?

Here are 5 related questions for further understanding:
1. What is the relationship between wavelength, frequency, and the speed of a wave?
2. How does the phase difference change if the distance between the two points increases?
3. Can phase differences be expressed in degrees instead of radians? How would you convert between them?
4. What happens to the wavelength if the speed of sound changes but the frequency remains constant?
5. What is the significance of a phase difference of \( 2\pi \) radians?

**Tip:** Phase difference can be converted from radians to degrees by multiplying by \( \frac{180}{\pi} \).

What is the phase difference between two points 0.16 m apart on a progressive sound wave of frequency 256 Hz, given the speed of sound is 330 m/s?

Learn how to calculate the phase difference between two points 0.16 m apart on a sound wave with a frequency of 256 Hz and a sound speed of 330 m/s. This solution covers key wave physics concepts, including wavelength and phase difference calculations.

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