Two sound waves one in air and the other in fresh water are equal in intensity.
(a) Find the ratio of pressure amplitude of the wave in water to that of the wave in air.
(b) If the pressure amplitudes of the waves are equal then what will be the ratio of the intensities of the waves.
[`V_(sound) = 340 m//s` in air & density of air `= 1.25 kg//m^(3), V_(sound) = 1530 m//s` in water, density of water `= 1000 kg//m^(3)`]

Answer

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Finding the ratio of pressure amplitudes

1. **Understanding the formula for intensity**:
The intensity \( I \) of a sound wave is given by the formula:
\[ I = \frac{P_m^2}{2 \rho V} \]
where \( P_m \) is the pressure amplitude, \( \rho \) is the density of the medium, and \( V \) is the velocity of sound in that medium.

2. **Setting up the equation for both media**:
Since the intensities of the sound waves in air and water are equal, we can write:
\[ I_w = I_a \]
This gives us:
\[ \frac{P_{w}^2}{2 \rho_{w} V_{w}} = \frac{P_{a}^2}{2 \rho_{a} V_{a}} \]

3. **Rearranging to find the ratio of pressure amplitudes**:
By cross-multiplying, we get:
\[ P_{w}^2 \cdot \rho_{a} \cdot V_{a} = P_{a}^2 \cdot \rho_{w} \cdot V_{w} \]
Taking the square root of both sides gives:
\[ \frac{P_{w}}{P_{a}} = \sqrt{\frac{\rho_{w} V_{w}}{\rho_{a} V_{a}}} \]

4. **Substituting the known values**:
- Density of water \( \rho_{w} = 1000 \, \text{kg/m}^3 \)
- Density of air \( \rho_{a} = 1.25 \, \text{kg/m}^3 \)
- Velocity of sound in water \( V_{w} = 1530 \, \text{m/s} \)
- Velocity of sound in air \( V_{a} = 340 \, \text{m/s} \)

Substituting these values into the equation:
\[ \frac{P_{w}}{P_{a}} = \sqrt{\frac{1000 \cdot 1530}{1.25 \cdot 340}} \]

5. **Calculating the ratio**:
\[ \frac{P_{w}}{P_{a}} = \sqrt{\frac{1000 \cdot 1530}{1.25 \cdot 340}} = \sqrt{\frac{1000 \cdot 1530}{425}} = \sqrt{3600} = 60 \]
Therefore, the ratio of the pressure amplitudes is:
\[ \frac{P_{w}}{P_{a}} = 60 \]

### Part (b): Finding the ratio of intensities

1. **Using the relationship between pressure amplitude and intensity**:
If the pressure amplitudes are equal, then we can use the formula for intensity:
\[ I = \frac{P_m^2}{2 \rho V} \]

2. **Setting up the ratio of intensities**:
If \( P_{w} = P_{a} \), then:
\[ \frac{I_{w}}{I_{a}} = \frac{\rho_{a} V_{a}}{\rho_{w} V_{w}} \]

3. **Substituting the known values**:
Using the same values as before:
\[ \frac{I_{w}}{I_{a}} = \frac{1.25 \cdot 340}{1000 \cdot 1530} \]

4. **Calculating the ratio**:
\[ \frac{I_{w}}{I_{a}} = \frac{425}{1530000} = \frac{1}{3600} \]
Therefore, the ratio of the intensities is:
\[ \frac{I_{w}}{I_{a}} = \frac{1}{3600} \]

### Final Answers:
(a) The ratio of pressure amplitudes \( \frac{P_{w}}{P_{a}} = 60 \)

(b) The ratio of intensities \( \frac{I_{w}}{I_{a}} = \frac{1}{3600} \)