A point source of sound is located inside sea water. Bulk modulus of sea water is `B_(omega)=2.0 xx 10^(9) N//m^(2)`. A diver located at a distance of `10` m from the source registers a pressure amplitude of `DeltaP_(0)=3000 pi N//m^(2)` and gives the equation of sound wave as
`y=A sin (15 pi x-21000 pit)`, y and x are in meter and `t` is in second.
(a) Find the displacement amplitude of the sound wave at the location of the diver.
(b) Find the power of the sound source.

Answer

Let's solve the problem step by step.

### Given Data:
- Bulk modulus of seawater, \( B = 2.0 \times 10^9 \, \text{N/m}^2 \)
- Pressure amplitude, \( \Delta P_0 = 3000 \pi \, \text{N/m}^2 \)
- Distance from the source to the diver, \( r = 10 \, \text{m} \)
- Wave equation: \( y = A \sin(15 \pi x - 21000 \pi t) \)

### Part (a): Find the displacement amplitude of the sound wave at the location of the diver.

1. **Identify the wave parameters:**
- From the wave equation, we can identify:
- Wave number \( k = 15 \pi \)
- Angular frequency \( \omega = 21000 \pi \)

2. **Calculate the wave velocity \( v \):**
\[ v = \frac{\omega}{k} = \frac{21000 \pi}{15 \pi} = 1400 \, \text{m/s} \]

3. **Calculate the density \( \rho \) of seawater:**
Using the formula \( \rho = \frac{B}{v^2} \):
\[ \rho = \frac{2.0 \times 10^9}{(1400)^2} = \frac{2.0 \times 10^9}{1960000} \approx 1020.41 \, \text{kg/m}^3 \]

4. **Relate pressure amplitude to displacement amplitude:**
The relationship between pressure amplitude and displacement amplitude is given by:
\[ \Delta P_0 = B \cdot k \cdot A \]
Rearranging for \( A \):
\[ A = \frac{\Delta P_0}{B \cdot k} \]

5. **Substituting the values:**
\[ A = \frac{3000 \pi}{(2.0 \times 10^9) \cdot (15 \pi)} \]
Simplifying:
\[ A = \frac{3000}{2.0 \times 10^9 \cdot 15} = \frac{3000}{30 \times 10^9} = \frac{1}{10^6} = 0.1 \, \mu m \]

### Part (b): Find the power of the sound source.

1. **Calculate the intensity \( I \):**
The intensity of the sound wave can be calculated using:
\[ I = \frac{\Delta P_0^2}{2 \rho v} \]
Substituting the values:
\[ I = \frac{(3000 \pi)^2}{2 \cdot (1020.41) \cdot (1400)} \]

2. **Calculating \( I \):**
\[ I = \frac{(3000^2 \cdot \pi^2)}{2 \cdot 1020.41 \cdot 1400} \]
\[ I \approx \frac{(9000000 \cdot \pi^2)}{2851454.8} \approx 31.07 \, \text{W/m}^2 \]

3. **Calculate the power \( P \):**
The power of the sound source is given by:
\[ P = I \cdot A \]
where \( A \) is the area over which the sound is spread, which is the surface area of a sphere:
\[ A = 4 \pi r^2 \]
Substituting \( r = 10 \, \text{m} \):
\[ A = 4 \pi (10)^2 = 400 \pi \, \text{m}^2 \]

4. **Calculating \( P \):**
\[ P = 31.07 \cdot 400 \pi \approx 31.07 \cdot 1256.64 \approx 39000 \, \text{W} \]

### Final Answers:
- (a) Displacement amplitude \( A \approx 0.1 \, \mu m \)
- (b) Power of the sound source \( P \approx 39000 \, \text{W} \)