If it were possible to generate a sinusoidal `300 H_(Z)` sound wave in air that has a displacement amplitude of `0.200 mm` . What would be the sound level ? (Assume `upsilon = 330 m//s` and `rho_(air) = 1.29 kg//m^(3))`

Explanation

To solve the problem of finding the sound level of a sinusoidal sound wave with a given frequency and displacement amplitude, we will follow these steps:

### Step 1: Write down the given values
- Frequency (f) = 300 Hz
- Displacement amplitude (A) = 0.200 mm = 0.200 × 10^(-3) m = 0.0002 m
- Velocity of sound in air (v) = 330 m/s
- Density of air (ρ) = 1.29 kg/m³

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 300 \approx 1884.96 \, \text{rad/s} \]

### Step 3: Calculate the intensity (I) of the sound wave
The intensity (I) of a sound wave can be calculated using the formula:
\[ I = \frac{1}{2} \rho v \omega^2 A^2 \]
Substituting the values we have:
\[ I = \frac{1}{2} \times 1.29 \, \text{kg/m}^3 \times 330 \, \text{m/s} \times (1884.96 \, \text{rad/s})^2 \times (0.0002 \, \text{m})^2 \]
Calculating this step by step:
1. Calculate \( \omega^2 \):
\[ \omega^2 \approx (1884.96)^2 \approx 3550000.57 \, \text{rad}^2/\text{s}^2 \]
2. Calculate \( A^2 \):
\[ A^2 = (0.0002)^2 = 4 \times 10^{-8} \, \text{m}^2 \]
3. Now plug these into the intensity formula:
\[ I \approx \frac{1}{2} \times 1.29 \times 330 \times 3550000.57 \times 4 \times 10^{-8} \]
\[ I \approx 30.27 \, \text{W/m}^2 \]

### Step 4: Calculate the sound level (L)
The sound level in decibels (dB) can be calculated using the formula:
\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \]
Where \( I_0 \) is the reference intensity, typically \( I_0 = 10^{-12} \, \text{W/m}^2 \).
Substituting the values:
\[ L = 10 \log_{10} \left(\frac{30.27}{10^{-12}}\right) \]
Calculating this:
\[ L = 10 \log_{10} (30.27 \times 10^{12}) = 10 \left(\log_{10} (30.27) + 12\right) \]
Using \( \log_{10} (30.27) \approx 1.48 \):
\[ L \approx 10 \times (1.48 + 12) = 10 \times 13.48 = 134.8 \, \text{dB} \]

### Final Answer
The sound level is approximately **134.8 dB**.

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