(a) A longitudinal wave propagating in a water-filled pipe has intensity `3.00xx10^(-6) W//m ^(2)` and frequency `3400 H_(Z)` . Find the amplitude `A` and wavelength `lambda` of the wave . Water has density `1000 kg//m ^(3)` and bulk modulus `2.18xx10^(9) Pa`. (b) If the pipe is filled with air at pressure `1.00 xx10^(5)` Pa and density `1.20 kg//m^(3)`, What will be the amplitude `A` and wavelength `lambda` of a longitudinal wave the same intensity and frequency as in part (a) ? (c ) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitude ? Why is this ratio so different from/ Conider air as diatomic.

Explanation

To solve the problem step by step, we will break it down into parts (a), (b), and (c) as given in the question.

### Part (a): Finding Amplitude \( A \) and Wavelength \( \lambda \) in Water

1. **Calculate the velocity of sound in water**:
The velocity of sound \( v \) in a medium can be calculated using the formula:
\[ v = \sqrt{\frac{K}{\rho}} \]
where \( K \) is the bulk modulus and \( \rho \) is the density of the medium. For water:
- \( K = 2.18 \times 10^9 \, \text{Pa} \)
- \( \rho = 1000 \, \text{kg/m}^3 \)

Substituting the values:
\[ v = \sqrt{\frac{2.18 \times 10^9}{1000}} = \sqrt{2.18 \times 10^6} \approx 1476 \, \text{m/s} \]

2. **Calculate the wavelength \( \lambda \)**:
The wavelength \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
where \( f \) is the frequency. Given \( f = 3400 \, \text{Hz} \):
\[ \lambda = \frac{1476}{3400} \approx 0.434 \, \text{m} \]

3. **Calculate the amplitude \( A \)**:
The intensity \( I \) of a wave is related to the amplitude by the formula:
\[ I = \frac{1}{2} \rho v \omega^2 A^2 \]
where \( \omega = 2 \pi f \). First, calculate \( \omega \):
\[ \omega = 2 \pi \times 3400 \approx 21363.0 \, \text{rad/s} \]
Rearranging the intensity formula to solve for \( A \):
\[ A = \sqrt{\frac{2I}{\rho v \omega^2}} \]
Substituting the values:
\[ A = \sqrt{\frac{2 \times 3.00 \times 10^{-6}}{1000 \times 1476 \times (21363.0)^2}} \]
After calculating, we find:
\[ A \approx 9.44 \times 10^{-11} \, \text{m} \]

### Part (b): Finding Amplitude \( A \) and Wavelength \( \lambda \) in Air

1. **Calculate the velocity of sound in air**:
The velocity of sound in air can be calculated using:
\[ v = \sqrt{\frac{\gamma P}{\rho}} \]
where \( \gamma = 1.4 \) (for diatomic gases), \( P = 1.00 \times 10^5 \, \text{Pa} \), and \( \rho = 1.20 \, \text{kg/m}^3 \):
\[ v = \sqrt{\frac{1.4 \times 10^5}{1.20}} \approx 341.56 \, \text{m/s} \]

2. **Calculate the wavelength \( \lambda \)**:
Using the same formula for wavelength:
\[ \lambda = \frac{v}{f} = \frac{341.56}{3400} \approx 0.100 \, \text{m} \]

3. **Calculate the amplitude \( A \)**:
Using the same intensity formula:
\[ A = \sqrt{\frac{2I}{\rho v \omega^2}} \]
Substituting the values for air:
\[ A = \sqrt{\frac{2 \times 3.00 \times 10^{-6}}{1.20 \times 341.56 \times (21363.0)^2}} \]
After calculating, we find:
\[ A \approx 5.6 \times 10^{-9} \, \text{m} \]

### Part (c): Comparing Amplitudes and Finding the Ratio

1. **Compare amplitudes**:
From the calculations:
- Amplitude in water: \( A_{water} \approx 9.44 \times 10^{-11} \, \text{m} \)
- Amplitude in air: \( A_{air} \approx 5.6 \times 10^{-9} \, \text{m} \)

Thus, the amplitude is larger in air.

2. **Calculate the ratio of amplitudes**:
\[ \text{Ratio} = \frac{A_{air}}{A_{water}} = \frac{5.6 \times 10^{-9}}{9.44 \times 10^{-11}} \approx 59.3 \approx 60 \]

3. **Explain the difference**:
The ratio is significantly different because the density of air is much lower than that of water, which affects how sound waves propagate through each medium. The intensity formula shows that the amplitude is inversely related to the square root of the density, leading to a much larger amplitude in air for the same intensity.