the maximum pressure variation that the human ear can tolerate in loud sound is about `30 N //m^(2) ` . The corresponding maximum displacement for a sound wave ina air having a frequency of ` 10^(3) Hz`is
take velocity of sound in air as 300 m/s and density of air ` 1.5 kg // m ^(3)`

Options

A. `(2pi)/3 xx10^(-2)m`
B. `(2 xx10 ^(-4))/pi m`
C. `(pi)/3 xx10^(-2)m`
D. `(10 ^(-4))/(3pi) m`

Answer

To find the maximum displacement (amplitude) of a sound wave in air given the maximum pressure variation, we can use the relationship between pressure variation, frequency, density, and amplitude of the wave.

### Step-by-step Solution:

1. **Identify the given values**:
- Maximum pressure variation, \( \Delta P = 30 \, \text{N/m}^2 \)
- Frequency, \( f = 10^3 \, \text{Hz} \)
- Velocity of sound in air, \( v = 300 \, \text{m/s} \)
- Density of air, \( \rho = 1.5 \, \text{kg/m}^3 \)

2. **Use the formula for maximum pressure variation**:
The relationship between maximum pressure variation (\( \Delta P \)), frequency (\( f \)), density (\( \rho \)), and amplitude (\( A \)) is given by:
\[ \Delta P = 2 \pi f \rho v A \]
Rearranging this formula to solve for amplitude \( A \):
\[ A = \frac{\Delta P}{2 \pi f \rho v} \]

3. **Substituting the values into the formula**:
\[ A = \frac{30}{2 \pi (10^3) (1.5) (300)} \]

4. **Calculating the denominator**:
- First, calculate \( 2 \pi \):
\[ 2 \pi \approx 6.2832 \]
- Now, calculate the entire denominator:
\[ 2 \pi (10^3) (1.5) (300) = 6.2832 \times 1000 \times 1.5 \times 300 \]
\[ = 6.2832 \times 450000 = 2827438.4 \]

5. **Calculating the amplitude**:
\[ A = \frac{30}{2827438.4} \approx 1.06 \times 10^{-5} \, \text{m} \]

6. **Final expression**:
To express this in a more manageable form:
\[ A \approx 1.06 \times 10^{-5} \, \text{m} \approx \frac{10^{-4}}{3\pi} \, \text{m} \]

### Conclusion:
The corresponding maximum displacement (amplitude) for the sound wave in air is approximately:
\[ A \approx \frac{10^{-4}}{3 \pi} \, \text{m} \]