The pressure at a point varies from `99980 Pa` to `100020 Pa` due to simple harmonic sound wave. The amplitude and wavelength of the wave are `5 xx 10^(-6) m` and `40 cm` respectively. Find the bulk modulus of air

Answer

To find the bulk modulus of air given the pressure variation due to a sound wave, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Maximum and Minimum Pressure:**
- The maximum pressure \( P_{max} = 100020 \, \text{Pa} \)
- The minimum pressure \( P_{min} = 99980 \, \text{Pa} \)

2. **Calculate the Pressure Amplitude \( P_0 \):**
- The pressure amplitude \( P_0 \) is given by:
\[ P_0 = \frac{P_{max} - P_{min}}{2} = \frac{100020 \, \text{Pa} - 99980 \, \text{Pa}}{2} = \frac{40 \, \text{Pa}}{2} = 20 \, \text{Pa} \]

3. **Convert the Wavelength to Meters:**
- Given wavelength \( \lambda = 40 \, \text{cm} \):
\[ \lambda = 40 \, \text{cm} = 40 \times 10^{-2} \, \text{m} = 0.4 \, \text{m} \]

4. **Identify the Displacement Amplitude \( A \):**
- The displacement amplitude \( A \) is given as:
\[ A = 5 \times 10^{-6} \, \text{m} \]

5. **Use the Formula for Bulk Modulus \( K \):**
- The bulk modulus \( K \) can be calculated using the formula:
\[ K = \frac{P_0 \cdot \lambda}{2 \pi A} \]
- Substitute the values:
\[ K = \frac{20 \, \text{Pa} \cdot 0.4 \, \text{m}}{2 \pi (5 \times 10^{-6} \, \text{m})} \]

6. **Calculate \( K \):**
- First, calculate the numerator:
\[ 20 \cdot 0.4 = 8 \, \text{Pa} \cdot \text{m} \]
- Now calculate the denominator:
\[ 2 \pi (5 \times 10^{-6}) \approx 3.14 \times 10^{-5} \, \text{m} \]
- Now substitute these into the equation:
\[ K = \frac{8}{3.14 \times 10^{-5}} \approx 2.55 \times 10^5 \, \text{N/m}^2 \]

### Final Answer:
The bulk modulus of air is approximately:
\[ K \approx 2.55 \times 10^5 \, \text{N/m}^2 \]