What is the amplitude of motion for the air in the path of a `60 dB` , `800 H_(Z)` sound wave? Assume that `rho_(air)= 1.29 kg//m^(3)` and `upsilon = 330 m//s`.

Explanation

To find the amplitude of motion for the air in the path of a 60 dB, 800 Hz sound wave, we can follow these steps:

### Step 1: Calculate the Intensity of the Sound Wave
We start with the formula for sound level in decibels (dB):
\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \]
where:
- \(L\) is the sound level in dB (60 dB),
- \(I\) is the intensity of the sound wave,
- \(I_0\) is the reference intensity, which is \(10^{-12} \, \text{W/m}^2\).

Rearranging the formula to solve for \(I\):
\[ 60 = 10 \log_{10} \left(\frac{I}{10^{-12}}\right) \]
Dividing both sides by 10:
\[ 6 = \log_{10} \left(\frac{I}{10^{-12}}\right) \]
Converting from logarithmic form:
\[ \frac{I}{10^{-12}} = 10^6 \]
Thus,
\[ I = 10^6 \times 10^{-12} = 10^{-6} \, \text{W/m}^2 \]

### Step 2: Relate Intensity to Amplitude
The intensity \(I\) of a sound wave can also be expressed in terms of the amplitude \(A\) as follows:
\[ I = \frac{1}{2} \rho v \omega^2 A^2 \]
where:
- \(\rho\) is the density of air (\(1.29 \, \text{kg/m}^3\)),
- \(v\) is the velocity of sound (\(330 \, \text{m/s}\)),
- \(\omega\) is the angular frequency (\(\omega = 2\pi f\)),
- \(A\) is the amplitude.

### Step 3: Calculate Angular Frequency
The angular frequency \(\omega\) can be calculated using the frequency \(f\):
\[ \omega = 2\pi f = 2\pi \times 800 \, \text{Hz} \approx 5026.55 \, \text{rad/s} \]

### Step 4: Substitute Values into the Intensity Formula
Now we substitute the values into the intensity formula:
\[ 10^{-6} = \frac{1}{2} \times 1.29 \times 330 \times (5026.55)^2 \times A^2 \]

### Step 5: Solve for Amplitude \(A\)
Rearranging the equation to solve for \(A^2\):
\[ A^2 = \frac{2 \times 10^{-6}}{1.29 \times 330 \times (5026.55)^2} \]
Calculating the denominator:
\[ 1.29 \times 330 \times (5026.55)^2 \approx 1.29 \times 330 \times 25265680.3 \approx 1.09 \times 10^{10} \]
Thus,
\[ A^2 = \frac{2 \times 10^{-6}}{1.09 \times 10^{10}} \approx 1.83 \times 10^{-17} \]
Taking the square root to find \(A\):
\[ A \approx \sqrt{1.83 \times 10^{-17}} \approx 1.36 \times 10^{-8} \, \text{m} \]
This can be expressed in nanometers:
\[ A \approx 13.6 \, \text{nm} \]

### Final Answer
The amplitude of the motion for the air in the path of the sound wave is approximately \(13.6 \, \text{nm}\).

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