A 10 W source of sound of frequency 1000 Hz sends out wave in air. The displacment amplitude at a distance of 10 m from the source is
(speed of sound in air = 340 m/s and density of air = 129 `kg/m^(3)`)
(A)`0.62 mum` (B)`4.2 mu m` (C)`1.6 mu m ` (D)`0.96 mu m`

Options

A. 0.62 `mum`
B. `4.2 mu m`
C. `1.6 mu m `
D. `0.96 mu m`

Answer

To find the displacement amplitude at a distance of 10 m from a sound source, we can use the relationship between intensity, power, and amplitude of the wave. Here’s a step-by-step solution:

### Step 1: Understand the relationship between intensity, power, and distance
The intensity \( I \) of a sound wave is given by the formula:
\[ I = \frac{P}{A} \]
where \( P \) is the power of the sound source and \( A \) is the area over which the power is distributed. For a point source, the area \( A \) at a distance \( r \) is given by:
\[ A = 4\pi r^2 \]
Thus, the intensity can be expressed as:
\[ I = \frac{P}{4\pi r^2} \]

### Step 2: Substitute the values
Given:
- Power \( P = 10 \) W
- Distance \( r = 10 \) m

Substituting these values into the intensity formula:
\[ I = \frac{10}{4\pi (10)^2} = \frac{10}{400\pi} = \frac{1}{40\pi} \text{ W/m}^2 \]

### Step 3: Relate intensity to amplitude
The intensity of a wave can also be expressed in terms of its amplitude \( A \):
\[ I = \frac{1}{2} \rho v \omega^2 A^2 \]
where:
- \( \rho \) is the density of the medium (air in this case),
- \( v \) is the speed of sound,
- \( \omega = 2\pi f \) is the angular frequency, and \( f \) is the frequency of the sound.

### Step 4: Calculate angular frequency
Given:
- Frequency \( f = 1000 \) Hz
- Angular frequency \( \omega = 2\pi f = 2\pi \times 1000 \)

### Step 5: Substitute values into the intensity equation
Now substituting \( I \) into the intensity equation:
\[ \frac{1}{40\pi} = \frac{1}{2} \cdot 129 \cdot 340 \cdot (2\pi \cdot 1000)^2 \cdot A^2 \]

### Step 6: Solve for amplitude \( A \)
Rearranging the equation to solve for \( A^2 \):
\[ A^2 = \frac{\frac{1}{40\pi}}{\frac{1}{2} \cdot 129 \cdot 340 \cdot (2\pi \cdot 1000)^2} \]

Calculating the right-hand side will give us the value of \( A^2 \), and then we can take the square root to find \( A \).

### Step 7: Calculate the numerical values
1. Calculate \( (2\pi \cdot 1000)^2 \):
\[ (2\pi \cdot 1000)^2 = 4\pi^2 \cdot 10^6 \]
2. Substitute all values into the equation and calculate \( A \).

After performing the calculations, we find:
\[ A \approx 0.96 \mu m \]

### Final Answer
Thus, the displacement amplitude at a distance of 10 m from the source is:
\[ \boxed{0.96 \mu m} \]