The intensity of a wave propagating through air is `0.22 Wm^(-2)`. If the frequency of the wave is 500 Hz calculate the displacement amplitude ? `["Density of air=1.29 kgm"^(3)`, speed of sound in air= 340 m/s]

Answer

To solve the problem, we will use the formula for the intensity of a wave, which is given by:

\[ I = 2 \pi^2 a^2 n^2 \rho v \]

Where:
- \( I \) = intensity of the wave (in W/m²)
- \( a \) = displacement amplitude (in meters)
- \( n \) = frequency of the wave (in Hz)
- \( \rho \) = density of the medium (in kg/m³)
- \( v \) = speed of sound in the medium (in m/s)

### Step 1: Identify the given values
From the question, we have:
- Intensity, \( I = 0.22 \, \text{W/m}^2 \)
- Frequency, \( n = 500 \, \text{Hz} \)
- Density of air, \( \rho = 1.29 \, \text{kg/m}^3 \)
- Speed of sound in air, \( v = 340 \, \text{m/s} \)

### Step 2: Rearrange the formula to solve for amplitude \( a \)
We need to isolate \( a \) in the intensity formula:

\[ a^2 = \frac{I}{2 \pi^2 n^2 \rho v} \]

Taking the square root will give us the displacement amplitude:

\[ a = \sqrt{\frac{I}{2 \pi^2 n^2 \rho v}} \]

### Step 3: Substitute the known values into the equation
Now we substitute the values into the rearranged formula:

\[ a = \sqrt{\frac{0.22}{2 \pi^2 (500)^2 (1.29)(340)}} \]

### Step 4: Calculate the values inside the square root
First, calculate \( 2 \pi^2 \):

\[ 2 \pi^2 \approx 2 \times (3.14)^2 \approx 19.63 \]

Now calculate \( (500)^2 \):

\[ (500)^2 = 250000 \]

Now calculate \( 19.63 \times 250000 \times 1.29 \times 340 \):

1. Calculate \( 19.63 \times 250000 = 4907500 \)
2. Calculate \( 4907500 \times 1.29 = 6328275 \)
3. Calculate \( 6328275 \times 340 \approx 2157603500 \)

Now, we have:

\[ a = \sqrt{\frac{0.22}{2157603500}} \]

### Step 5: Calculate the final value of \( a \)
Now calculate the fraction:

\[ \frac{0.22}{2157603500} \approx 1.019 \times 10^{-10} \]

Taking the square root:

\[ a \approx \sqrt{1.019 \times 10^{-10}} \approx 1.01 \times 10^{-5} \, \text{m} \]

### Final Answer
Thus, the displacement amplitude \( a \) is approximately:

\[ a \approx 1.01 \times 10^{-5} \, \text{m} \]