A sound wave in air has a frequency of `300 H_(Z)` and a displacement ampulitude of `6.0 xx10^(-3) mm`. For this sound waves calculate the (a) Pressure ampulitude (b) intensity (c ) Sound intensity level (in dB)
Speed of sound `= 344 m//s` and density of air `= 1.2 kg//m^(3)`.

Explanation

To solve the problem step by step, we will calculate the pressure amplitude, intensity, and sound intensity level of the sound wave given the frequency, displacement amplitude, speed of sound, and density of air.

### Given Data:
- Frequency, \( f = 300 \, \text{Hz} \)
- Displacement Amplitude, \( A = 6.0 \times 10^{-3} \, \text{mm} = 6.0 \times 10^{-6} \, \text{m} \) (conversion from mm to m)
- Speed of Sound, \( v = 344 \, \text{m/s} \)
- Density of Air, \( \rho = 1.2 \, \text{kg/m}^3 \)

### (a) Calculate Pressure Amplitude

The pressure amplitude \( \Delta P \) can be calculated using the formula:
\[ \Delta P = \rho v \omega A \]
where \( \omega = 2 \pi f \) is the angular frequency.

1. **Calculate \( \omega \)**:
\[ \omega = 2 \pi f = 2 \pi (300) = 600 \pi \approx 1884.96 \, \text{rad/s} \]

2. **Substitute values into the pressure amplitude formula**:
\[ \Delta P = \rho v \omega A = (1.2 \, \text{kg/m}^3)(344 \, \text{m/s})(1884.96 \, \text{rad/s})(6.0 \times 10^{-6} \, \text{m}) \]

3. **Calculate**:
\[ \Delta P \approx 1.2 \times 344 \times 1884.96 \times 6.0 \times 10^{-6} \approx 0.04666 \, \text{Pa} \]

### (b) Calculate Intensity

The intensity \( I \) of the sound wave can be calculated using the formula:
\[ I = \frac{1}{2} \rho v \omega^2 A^2 \]

1. **Substitute values into the intensity formula**:
\[ I = \frac{1}{2} (1.2 \, \text{kg/m}^3)(344 \, \text{m/s})(1884.96 \, \text{rad/s})^2(6.0 \times 10^{-6} \, \text{m})^2 \]

2. **Calculate**:
\[ I \approx \frac{1}{2} \times 1.2 \times 344 \times (1884.96)^2 \times (6.0 \times 10^{-6})^2 \approx 0.0264 \, \text{W/m}^2 \]

### (c) Calculate Sound Intensity Level in dB

The sound intensity level \( L \) in decibels (dB) can be calculated using the formula:
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where \( I_0 = 10^{-12} \, \text{W/m}^2 \) is the reference intensity.

1. **Substitute values into the sound intensity level formula**:
\[ L = 10 \log_{10} \left( \frac{0.0264}{10^{-12}} \right) \]

2. **Calculate**:
\[ L \approx 10 \log_{10} (2.64 \times 10^{10}) \approx 10 \times (10 + \log_{10}(2.64)) \approx 10 \times (10 + 0.423) \approx 104.23 \, \text{dB} \]

### Final Answers:
- (a) Pressure Amplitude: \( \Delta P \approx 0.04666 \, \text{Pa} \)
- (b) Intensity: \( I \approx 0.0264 \, \text{W/m}^2 \)
- (c) Sound Intensity Level: \( L \approx 104.23 \, \text{dB} \)