To solve the problem of calculating the amplitude of the sound wave at a distance of 25 m from the point source, we can follow these steps:
### Step 1: Understand the Given Information
- Frequency of the sound wave, \( f = 256 \, \text{Hz} \)
- Power of the source, \( P = 5 \, \text{W} \) (since 1 Joule/second = 1 Watt)
- Distance from the source, \( R = 25 \, \text{m} \)
- Speed of sound in air, \( c = 330 \, \text{m/s} \)
- Density of air, \( \rho = 1.29 \, \text{kg/m}^3 \)
### Step 2: Calculate the Intensity of the Sound Wave
The intensity \( I \) of the sound wave at a distance \( R \) from the source can be calculated using the formula:
\[
I = \frac{P}{4 \pi R^2}
\]
Substituting the values:
\[
I = \frac{5}{4 \pi (25)^2}
\]
Calculating \( 25^2 = 625 \):
\[
I = \frac{5}{4 \pi \times 625}
\]
\[
I = \frac{5}{2500 \pi}
\]
\[
I \approx \frac{5}{7850} \approx 0.0006366 \, \text{W/m}^2
\]
### Step 3: Relate Intensity to Amplitude
The intensity of a sound wave is also related to its amplitude \( A \) by the formula:
\[
I = \frac{1}{2} \rho c \omega^2 A^2
\]
where \( \omega = 2 \pi f \) is the angular frequency. First, we calculate \( \omega \):
\[
\omega = 2 \pi f = 2 \pi \times 256 \approx 1608.5 \, \text{rad/s}
\]
### Step 4: Substitute and Solve for Amplitude
Now, we can substitute \( I \), \( \rho \), \( c \), and \( \omega \) into the intensity formula to solve for \( A \):
\[
0.0006366 = \frac{1}{2} \times 1.29 \times 330 \times (1608.5)^2 \times A^2
\]
Calculating the right side:
\[
\frac{1}{2} \times 1.29 \times 330 \approx 212.85
\]
\[
(1608.5)^2 \approx 2587280.25
\]
Thus,
\[
0.0006366 = 212.85 \times 2587280.25 \times A^2
\]
Calculating \( 212.85 \times 2587280.25 \):
\[
= 550000000 \, \text{(approximately)}
\]
Now,
\[
0.0006366 = 550000000 \times A^2
\]
\[
A^2 = \frac{0.0006366}{550000000}
\]
\[
A^2 \approx 1.157 \times 10^{-12}
\]
Taking the square root:
\[
A \approx \sqrt{1.157 \times 10^{-12}} \approx 1.077 \times 10^{-6} \, \text{m}
\]
### Final Answer
The amplitude of the wave at a distance of 25 m from the source is approximately:
\[
A \approx 10^{-6} \, \text{m} \text{ or } 1 \, \mu m
\]
