A point sound source is situated in a medium of dulk modulus `1.6 xx 10^(5) N//m^(2)`. An observer standing at a distance `10 m `from the source writes down the equeation for the wave as `y = A sin (15 pi x - 6000 pi t)`. Here `y` and `x` are in meter and `t` is in second. The maximum pressure ampulitude received to the observer's ear is `(24 pi)` pa, then find.
(a) the density of the medium,
(b) the displacement ampulitude `A` of the wave recived by the observer and
(c ) the power of the sound source.

Explanation

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

### Given Data:
- Bulk modulus \( B = 1.6 \times 10^5 \, \text{N/m}^2 \)
- Distance from the source \( r = 10 \, \text{m} \)
- Wave equation: \( y = A \sin(15\pi x - 6000\pi t) \)
- Maximum pressure amplitude \( P_0 = 24\pi \, \text{Pa} \)

### Step 1: Find the Density of the Medium
The relationship between the speed of sound \( v \), bulk modulus \( B \), and density \( \rho \) is given by:
\[ v = \sqrt{\frac{B}{\rho}} \]
First, we need to find the speed of sound \( v \) from the wave equation:
- The wave number \( k = 15\pi \)
- The angular frequency \( \omega = 6000\pi \)

The speed of sound \( v \) can be calculated as:
\[ v = \frac{\omega}{k} = \frac{6000\pi}{15\pi} = 400 \, \text{m/s} \]

Now, substituting \( v \) into the density formula:
\[ \rho = \frac{B}{v^2} = \frac{1.6 \times 10^5}{(400)^2} \]
Calculating \( (400)^2 = 160000 \):
\[ \rho = \frac{1.6 \times 10^5}{1.6 \times 10^5} = 1 \, \text{kg/m}^3 \]

### Step 2: Find the Displacement Amplitude \( A \)
The relationship between the pressure amplitude \( P_0 \), bulk modulus \( B \), displacement amplitude \( A \), and wave number \( k \) is given by:
\[ P_0 = B \cdot A \cdot k \]
Rearranging to find \( A \):
\[ A = \frac{P_0}{B \cdot k} \]
Substituting the values:
\[ A = \frac{24\pi}{1.6 \times 10^5 \cdot 15\pi} \]
Simplifying:
\[ A = \frac{24}{1.6 \times 10^5 \cdot 15} = \frac{24}{2.4 \times 10^6} = 10^{-5} \, \text{m} \]

### Step 3: Find the Power of the Sound Source
The intensity \( I \) of the sound wave can be expressed as:
\[ I = \frac{P}{A} \]
Where \( A \) is the area over which the power is distributed:
\[ A = 4\pi r^2 \]
The intensity can also be expressed in terms of the medium's properties:
\[ I = \frac{1}{2} \rho v \omega^2 A^2 \]
Setting the two expressions for intensity equal gives:
\[ \frac{P}{4\pi r^2} = \frac{1}{2} \rho v \omega^2 A^2 \]
Rearranging for \( P \):
\[ P = 4\pi r^2 \cdot \frac{1}{2} \rho v \omega^2 A^2 \]
Substituting the known values:
\[ P = 4\pi (10)^2 \cdot \frac{1}{2} \cdot 1 \cdot 400 \cdot (6000\pi)^2 \cdot (10^{-5})^2 \]
Calculating:
\[ P = 4\pi (100) \cdot 0.5 \cdot 1 \cdot 400 \cdot 36 \times 10^{10} \cdot 10^{-10} \]
\[ P = 200\pi \cdot 400 \cdot 36 \approx 288\pi^3 \, \text{W} \]

### Final Answers:
(a) Density of the medium: \( \rho = 1 \, \text{kg/m}^3 \)
(b) Displacement amplitude \( A = 10^{-5} \, \text{m} \)
(c) Power of the sound source \( P \approx 288\pi^3 \, \text{W} \)