The faintest sound the human ear can detect at a frequency of 1kHz (for which the ear is most sensitive) corresponds to an intensity of about `10^(-12)W//m^2` (the so called threshold of hearing). Determine the pressure amplitude and maximum displacement associated with this sound assuming the density of air = `1.3kg//m^2` and velocity of sound in air = 332 m/s

Options

A. `2.94 xx 10^(-5) N//m^2 , 1.1 xx 10^(-11) m`
B. `2.94 xx 10^(-4) N//m^2 , 1.3 xx 10^(-11) m`
C. `2.94 xx 10^(-1) N//m^2 , 1.2 xx 10^(-11) m`
D. `2.94 xx 10^(-3) N//m^2 , 1.5 xx 10^(-11) m`

Answer

To solve the problem, we need to determine the pressure amplitude (P) and the maximum displacement (A) associated with the sound intensity given. We will use the formulas related to sound intensity, pressure amplitude, and displacement amplitude.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Intensity of sound, \( I = 10^{-12} \, \text{W/m}^2 \)
- Density of air, \( \rho = 1.3 \, \text{kg/m}^3 \)
- Velocity of sound in air, \( V = 332 \, \text{m/s} \)

2. **Use the Formula for Intensity:**
The intensity \( I \) of a sound wave is related to the pressure amplitude \( P \) by the formula:
\[ I = \frac{P^2}{2 \rho V} \]
Rearranging this formula to solve for \( P \):
\[ P = \sqrt{2 I \rho V} \]

3. **Substitute the Known Values:**
Plugging in the values:
\[ P = \sqrt{2 \times 10^{-12} \, \text{W/m}^2 \times 1.3 \, \text{kg/m}^3 \times 332 \, \text{m/s}} \]

4. **Calculate the Pressure Amplitude \( P \):**
- First, calculate the product inside the square root:
\[ 2 \times 10^{-12} \times 1.3 \times 332 = 8.632 \times 10^{-10} \]
- Now take the square root:
\[ P = \sqrt{8.632 \times 10^{-10}} \approx 2.94 \times 10^{-5} \, \text{N/m}^2 \]

5. **Determine the Angular Frequency \( \omega \):**
The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi f \]
where \( f = 1 \, \text{kHz} = 10^3 \, \text{Hz} \):
\[ \omega = 2 \pi \times 10^3 \approx 6283.19 \, \text{rad/s} \]

6. **Use the Formula for Maximum Displacement Amplitude \( A \):**
The relationship between pressure amplitude \( P \), density \( \rho \), velocity \( V \), and displacement amplitude \( A \) is given by:
\[ P = \rho V \omega A \]
Rearranging to solve for \( A \):
\[ A = \frac{P}{\rho V \omega} \]

7. **Substitute the Known Values to Find \( A \):**
\[ A = \frac{2.94 \times 10^{-5}}{1.3 \times 332 \times 6283.19} \]

8. **Calculate the Maximum Displacement \( A \):**
- First, calculate the denominator:
\[ 1.3 \times 332 \times 6283.19 \approx 2.754 \times 10^3 \]
- Now calculate \( A \):
\[ A \approx \frac{2.94 \times 10^{-5}}{2.754 \times 10^3} \approx 1.07 \times 10^{-11} \, \text{m} \]

### Final Results:
- Pressure Amplitude \( P \approx 2.94 \times 10^{-5} \, \text{N/m}^2 \)
- Maximum Displacement \( A \approx 1.07 \times 10^{-11} \, \text{m} \)