For a person with normal hearing, the faintest sound that can be at a frequency of `400 Hz` has pressure amplitude of about `6.0 xx 10^(-5)` Pa . Calculate the corresponding intensity in `W//m^(2)` . Take speed of sound in air as `344 m//s` and density of air `1.2 kg//m^(3)` .

Explanation

To calculate the intensity of a sound wave given the pressure amplitude, we can follow these steps:

### Step 1: Identify the given values
- Pressure amplitude, \( \Delta P_m = 6.0 \times 10^{-5} \, \text{Pa} \)
- Speed of sound in air, \( V = 344 \, \text{m/s} \)
- Density of air, \( \rho = 1.2 \, \text{kg/m}^3 \)

### Step 2: Use the formula for intensity
The intensity \( I \) of a sound wave can be calculated using the formula:
\[ I = \frac{\Delta P_m^2}{2 \rho V} \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ I = \frac{(6.0 \times 10^{-5})^2}{2 \times 1.2 \times 344} \]

### Step 4: Calculate \( \Delta P_m^2 \)
First, calculate \( \Delta P_m^2 \):
\[ (6.0 \times 10^{-5})^2 = 3.6 \times 10^{-9} \, \text{Pa}^2 \]

### Step 5: Calculate the denominator
Now calculate the denominator:
\[ 2 \times 1.2 \times 344 = 825.6 \, \text{kg/(m}^2\text{s}^2) \]

### Step 6: Calculate the intensity
Now substitute back into the intensity formula:
\[ I = \frac{3.6 \times 10^{-9}}{825.6} \]

### Step 7: Perform the final calculation
Calculating the intensity:
\[ I \approx 4.36 \times 10^{-12} \, \text{W/m}^2 \]

### Final Result
Thus, the corresponding intensity is approximately:
\[ I \approx 4.4 \times 10^{-12} \, \text{W/m}^2 \]

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