The speed of a certain compressional wave in air at standard temperature and pressure is `330 m//s` . A point source of frequency `300 H_(Z)` radiates energy uniformly in all directions at the rate of `5` Watt. (a) What is the intensity of the wave at a distance of `20 m` from the source? (b) What is the amplitude of the wave there? [ Density of air at `STP = 1.29 kg //m^(3)`]

Explanation

To solve the problem step by step, let's break it down into two parts: (a) calculating the intensity of the wave at a distance of 20 m from the source, and (b) calculating the amplitude of the wave.

### Part (a): Calculate the Intensity

1. **Identify the Given Values**:
- Power of the source, \( P = 5 \, \text{W} \)
- Distance from the source, \( r = 20 \, \text{m} \)
- Formula for intensity, \( I = \frac{P}{A} \)
- Area for a point source radiating uniformly, \( A = 4\pi r^2 \)

2. **Calculate the Area**:
\[ A = 4\pi r^2 = 4\pi (20)^2 = 4 \times 3.14 \times 400 = 5024 \, \text{m}^2 \]

3. **Calculate the Intensity**:
\[ I = \frac{P}{A} = \frac{5}{5024} \approx 9.95 \times 10^{-4} \, \text{W/m}^2 \]

### Part (b): Calculate the Amplitude

1. **Identify the Given Values**:
- Intensity calculated, \( I \approx 9.95 \times 10^{-4} \, \text{W/m}^2 \)
- Frequency, \( f = 300 \, \text{Hz} \)
- Density of air, \( \rho = 1.29 \, \text{kg/m}^3 \)
- Speed of sound in air, \( v = 330 \, \text{m/s} \)

2. **Use the Formula for Intensity**:
\[ I = \frac{1}{2} \rho v (2\pi f A)^2 \]
Rearranging for \( A \):
\[ A^2 = \frac{I}{\frac{1}{2} \rho v (2\pi f)^2} \]

3. **Substitute the Values**:
\[ A^2 = \frac{9.95 \times 10^{-4}}{\frac{1}{2} \times 1.29 \times 330 \times (2\pi \times 300)^2} \]

4. **Calculate the Denominator**:
- Calculate \( 2\pi f \):
\[ 2\pi f = 2 \times 3.14 \times 300 \approx 1884 \, \text{rad/s} \]
- Calculate \( (2\pi f)^2 \):
\[ (2\pi f)^2 \approx 1884^2 \approx 3554256 \, \text{(rad/s)}^2 \]
- Calculate the entire denominator:
\[ \frac{1}{2} \times 1.29 \times 330 \times 3554256 \approx 7.5 \times 10^6 \]

5. **Final Calculation for Amplitude**:
\[ A^2 \approx \frac{9.95 \times 10^{-4}}{7.5 \times 10^6} \approx 1.31 \times 10^{-12} \]
\[ A \approx \sqrt{1.31 \times 10^{-12}} \approx 1.15 \times 10^{-6} \, \text{m} = 1150 \, \text{nm} \]

### Final Answers:
- (a) Intensity at 20 m: \( I \approx 9.95 \times 10^{-4} \, \text{W/m}^2 \)
- (b) Amplitude at 20 m: \( A \approx 1150 \, \text{nm} \)