The faintest sound the human ear can detect at a frequency of ` kHz` (for which ear is most sensitive) corresponds to an intensity of about `10^(-12)w//m^(2)`. Assuming the density of air `cong1.5kg//m^(3)` and velocity of sound in air `cong300m//s`, the pressure amplitude and displacement amplitude of the sound will be rspectively ____`N//m^(2)` and ____`m`.

Options

A. `3xx10^(-5) Pa`
B. `2xx10^(-5) Pa`
C. `5xx10^(-5) Pa`
D. `4xx10^(-5) Pa`

Explanation

To solve the problem, we will find the pressure amplitude and displacement amplitude of the sound wave step by step.

### Step 1: Given Data
- Frequency, \( f = 1 \, \text{kHz} = 1000 \, \text{Hz} \)
- Intensity, \( I = 10^{-12} \, \text{W/m}^2 \)
- Density of air, \( \rho = 1.5 \, \text{kg/m}^3 \)
- Velocity of sound in air, \( v = 300 \, \text{m/s} \)

### Step 2: Calculate Pressure Amplitude
We know the relationship between intensity and pressure amplitude is given by:

\[ I = \frac{P^2}{2 \rho v} \]

Where:
- \( I \) is the intensity,
- \( P \) is the pressure amplitude,
- \( \rho \) is the density,
- \( v \) is the velocity of sound.

Rearranging the formula to solve for pressure amplitude \( P \):

\[ P = \sqrt{2 I \rho v} \]

Now, substituting the known values into the equation:

\[ P = \sqrt{2 \times 10^{-12} \, \text{W/m}^2 \times 1.5 \, \text{kg/m}^3 \times 300 \, \text{m/s}} \]

Calculating the right-hand side:

\[ P = \sqrt{2 \times 10^{-12} \times 1.5 \times 300} \]
\[ P = \sqrt{2 \times 10^{-12} \times 450} \]
\[ P = \sqrt{9 \times 10^{-10}} = 3 \times 10^{-5} \, \text{N/m}^2 \]

### Step 3: Calculate Displacement Amplitude
The relationship between pressure amplitude and displacement amplitude is given by:

\[ P = \rho v \omega A \]

Where:
- \( A \) is the displacement amplitude,
- \( \omega = 2 \pi f \) is the angular frequency.

Rearranging the formula to solve for displacement amplitude \( A \):

\[ A = \frac{P}{\rho v \omega} \]

Substituting \( \omega = 2 \pi f \):

\[ A = \frac{P}{\rho v (2 \pi f)} \]

Now substituting the known values:

\[ A = \frac{3 \times 10^{-5}}{1.5 \times 300 \times 2 \pi \times 1000} \]

Calculating the denominator:

\[ A = \frac{3 \times 10^{-5}}{1.5 \times 300 \times 2000 \pi} \]
\[ = \frac{3 \times 10^{-5}}{900000 \pi} \]

Calculating \( A \):

\[ A = \frac{3 \times 10^{-5}}{9 \times 10^{5} \pi} = \frac{1}{3 \pi} \times 10^{-10} \, \text{m} \]

### Final Answers
- Pressure Amplitude, \( P = 3 \times 10^{-5} \, \text{N/m}^2 \)
- Displacement Amplitude, \( A = \frac{1}{3 \pi} \times 10^{-10} \, \text{m} \)