A point source of power 50W is producing sound waves of frequency 1875 Hz. The velocity of sound is 330 m/s, atmospheric pressure is 1.0×10 5 Nm −2 , density of air is 1.0 kgm −3 . Then pressure amplitude at r= 330 ​ m from the point source is (usingπ=22/7):

Answer

To find the pressure amplitude at a distance \( r = 330 \, \text{m} \) from a point source of sound, we can follow these steps:

### Step 1: Understand the relationship between power, intensity, and pressure amplitude
The intensity \( I \) of a sound wave can be expressed in terms of the pressure amplitude \( p_0 \):
\[ I = \frac{p_0^2}{2 \rho v} \]
where:
- \( I \) is the intensity,
- \( p_0 \) is the pressure amplitude,
- \( \rho \) is the density of air,
- \( v \) is the velocity of sound.

### Step 2: Calculate the intensity at a distance \( r \)
The intensity \( I \) at a distance \( r \) from a point source is also given by:
\[ I = \frac{P}{4 \pi r^2} \]
where \( P \) is the power of the source.

### Step 3: Substitute the values
Given:
- Power \( P = 50 \, \text{W} \)
- Distance \( r = 330 \, \text{m} \)
- Density \( \rho = 1.0 \, \text{kg/m}^3 \)
- Velocity \( v = 330 \, \text{m/s} \)

First, calculate the intensity \( I \):
\[ I = \frac{50}{4 \pi (330)^2} \]

### Step 4: Calculate \( 4 \pi (330)^2 \)
Calculating \( 4 \pi (330)^2 \):
\[ 4 \pi (330)^2 = 4 \times \frac{22}{7} \times 108900 = \frac{88}{7} \times 108900 \]
Calculating this gives:
\[ \approx 88 \times 15557.14 \approx 1375 \times 88 \approx 1375 \times 88 \approx 121000 \]

### Step 5: Calculate the intensity \( I \)
Now substituting back:
\[ I = \frac{50}{121000} \approx 0.0004132 \, \text{W/m}^2 \]

### Step 6: Relate intensity to pressure amplitude
Now, we can relate this intensity back to the pressure amplitude using the formula:
\[ I = \frac{p_0^2}{2 \rho v} \]
Rearranging gives:
\[ p_0^2 = 2 I \rho v \]
Substituting the values:
\[ p_0^2 = 2 \times 0.0004132 \times 1.0 \times 330 \]
Calculating this gives:
\[ p_0^2 \approx 0.272 \implies p_0 \approx \sqrt{0.272} \approx 0.52 \, \text{Pa} \]

### Step 7: Final result
Thus, the pressure amplitude at \( r = 330 \, \text{m} \) from the point source is approximately:
\[ p_0 \approx 0.52 \, \text{Pa} \]