A blast given a sound of intensity `0.8 W//m^(2)` at frequency 1kHz. If the denstiy of air is 1.3 `kg//m^(3)` and speed of sound in air is 330 m/s, then the amplitude of the sound wave is approximately

Options

A. `5 xx 10^(-6)m`
B. `15 xx 10^(-6)m`
C. `9.7 xx 10^(-6)m`
D. `20 xx 10^(-6)m`

Explanation

To find the amplitude of the sound wave given the intensity, frequency, density of air, and speed of sound, we can follow these steps:

### Step 1: Understand the relationship between intensity, amplitude, density, and frequency.
The intensity \( I \) of a sound wave can be expressed in terms of its amplitude \( A \), density \( \rho \), and angular frequency \( \omega \) using the formula:
\[ I = \frac{1}{2} \rho v A^2 \]
where:
- \( I \) is the intensity in \( W/m^2 \),
- \( \rho \) is the density of the medium (air in this case) in \( kg/m^3 \),
- \( v \) is the speed of sound in the medium in \( m/s \),
- \( A \) is the amplitude in meters.

### Step 2: Rearrange the formula to solve for amplitude \( A \).
From the intensity formula, we can rearrange it to solve for \( A \):
\[ A^2 = \frac{2I}{\rho v} \]
Taking the square root gives:
\[ A = \sqrt{\frac{2I}{\rho v}} \]

### Step 3: Substitute the known values into the formula.
Given:
- Intensity \( I = 0.8 \, W/m^2 \)
- Density \( \rho = 1.3 \, kg/m^3 \)
- Speed of sound \( v = 330 \, m/s \)

Substituting these values into the amplitude formula:
\[ A = \sqrt{\frac{2 \times 0.8}{1.3 \times 330}} \]

### Step 4: Calculate the values inside the square root.
Calculating the numerator:
\[ 2 \times 0.8 = 1.6 \]
Calculating the denominator:
\[ 1.3 \times 330 = 429 \]
Now substituting these results back into the amplitude formula:
\[ A = \sqrt{\frac{1.6}{429}} \]

### Step 5: Perform the final calculation.
Calculating the fraction:
\[ \frac{1.6}{429} \approx 0.00373 \]
Now, taking the square root:
\[ A \approx \sqrt{0.00373} \approx 0.061 \]
Converting to scientific notation:
\[ A \approx 6.1 \times 10^{-2} \, m \]

### Final Result:
The amplitude of the sound wave is approximately:
\[ A \approx 6.1 \times 10^{-2} \, m \]