The audible frequency for a normal human being is `20 Hz` to `20 kHz`. Fuind the corresponding wavelengths if the speed of sound in air `320 m//s`.
(b) Find the phase difference between two position separated by `20 cm` at a particular instant.

Explanation

To solve the problem step by step, we will break it down into two parts: finding the wavelengths corresponding to the given frequencies and then calculating the phase difference between two positions separated by a specified distance.

### Part (a): Finding the Wavelengths

1. **Identify the Given Values**:
- Speed of sound in air, \( v = 320 \, \text{m/s} \)
- Minimum audible frequency, \( f_{\text{min}} = 20 \, \text{Hz} \)
- Maximum audible frequency, \( f_{\text{max}} = 20 \, \text{kHz} = 20 \times 10^3 \, \text{Hz} \)

2. **Use the Wavelength Formula**:
The wavelength \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]

3. **Calculate Wavelength for Minimum Frequency**:
\[ \lambda_1 = \frac{v}{f_{\text{min}}} = \frac{320 \, \text{m/s}}{20 \, \text{Hz}} = 16 \, \text{m} \]

4. **Calculate Wavelength for Maximum Frequency**:
\[ \lambda_2 = \frac{v}{f_{\text{max}}} = \frac{320 \, \text{m/s}}{20 \times 10^3 \, \text{Hz}} = \frac{320}{20000} = 0.016 \, \text{m} = 16 \times 10^{-3} \, \text{m} \]

### Summary of Part (a):
- Wavelength for 20 Hz: \( \lambda_1 = 16 \, \text{m} \)
- Wavelength for 20 kHz: \( \lambda_2 = 16 \times 10^{-3} \, \text{m} \)

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### Part (b): Finding the Phase Difference

1. **Identify the Given Values**:
- Path difference, \( \Delta x = 20 \, \text{cm} = 0.2 \, \text{m} \)

2. **Use the Phase Difference Formula**:
The phase difference \( \Delta \phi \) can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]

3. **Calculate Phase Difference for \( \lambda_1 \)**:
\[ \Delta \phi_1 = \frac{2\pi}{\lambda_1} \Delta x = \frac{2\pi}{16} \times 0.2 = \frac{\pi}{40} \]

4. **Calculate Phase Difference for \( \lambda_2 \)**:
\[ \Delta \phi_2 = \frac{2\pi}{\lambda_2} \Delta x = \frac{2\pi}{16 \times 10^{-3}} \times 0.2 = \frac{2\pi \times 0.2}{16 \times 10^{-3}} = \frac{0.4\pi}{16 \times 10^{-3}} = 25\pi \]

### Summary of Part (b):
- Phase difference for \( \lambda_1 \): \( \Delta \phi_1 = \frac{\pi}{40} \)
- Phase difference for \( \lambda_2 \): \( \Delta \phi_2 = 25\pi \)

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