Plane harmonic waves of frequency 500 Hz are produced in air with displacement amplitude of `10mum`. Given that density of air is `1.29(kg)/(m^3)` and speed of sound in air is `340(m)/(s)`. Then

Options

A. the pressure amplitude is `13.8(N)/(m^2)`
B. the energy density is `6.4xx10^-4(J)/(m^3)`
C. the energy flux is `0.22(J)/((m^2s))`
D. only (a) and (c ) are correct

Explanation

To solve the given problem step by step, we will calculate the bulk modulus, wavelength, wave number, pressure, and energy density of the sound wave produced in air.

### Step 1: Calculate the Bulk Modulus (B)
The formula for the speed of sound (V) in a medium is given by:
\[ V = \sqrt{\frac{B}{\rho}} \]
Where:
- \( V \) = speed of sound = 340 m/s
- \( B \) = bulk modulus
- \( \rho \) = density of air = 1.29 kg/m³

Rearranging the formula to find the bulk modulus:
\[ B = V^2 \cdot \rho \]

Now substituting the values:
\[ B = (340)^2 \cdot (1.29) \]
\[ B = 115600 \cdot 1.29 \]
\[ B = 149124 \, \text{Pa} \]

### Step 2: Calculate the Wavelength (λ)
The relationship between speed (V), frequency (f), and wavelength (λ) is given by:
\[ V = f \cdot \lambda \]
Where:
- \( f \) = frequency = 500 Hz

Rearranging the formula to find the wavelength:
\[ \lambda = \frac{V}{f} \]

Substituting the values:
\[ \lambda = \frac{340}{500} \]
\[ \lambda = 0.68 \, \text{m} \]

### Step 3: Calculate the Wave Number (k)
The wave number (k) is defined as:
\[ k = \frac{2\pi}{\lambda} \]

Substituting the value of λ:
\[ k = \frac{2\pi}{0.68} \]
\[ k \approx 9.24 \, \text{rad/m} \]

### Step 4: Calculate the Pressure Amplitude (P)
The pressure amplitude (P) of a sound wave can be calculated using the formula:
\[ P = B \cdot A \cdot k \]
Where:
- \( A \) = displacement amplitude = \( 10 \, \mu m = 10 \times 10^{-6} \, m \)

Substituting the values:
\[ P = 149124 \cdot (10 \times 10^{-6}) \cdot 9.24 \]
\[ P \approx 13.8 \, \text{N/m}^2 \]

### Step 5: Calculate the Energy Density (U)
The energy density (U) of a sound wave is given by:
\[ U = \frac{1}{2} \cdot \rho \cdot \omega^2 \cdot A^2 \]
Where:
- \( \omega = 2\pi f \)

Calculating \( \omega \):
\[ \omega = 2\pi \cdot 500 \approx 3141.59 \, \text{rad/s} \]

Now substituting the values into the energy density formula:
\[ U = \frac{1}{2} \cdot 1.29 \cdot (3141.59)^2 \cdot (10 \times 10^{-6})^2 \]
\[ U \approx \frac{1}{2} \cdot 1.29 \cdot 9870.8 \cdot 10^{-12} \]
\[ U \approx 6.4 \times 10^{-4} \, \text{J/m}^3 \]

### Summary of Results
- Bulk Modulus (B): 149124 Pa
- Wavelength (λ): 0.68 m
- Wave Number (k): 9.24 rad/m
- Pressure Amplitude (P): 13.8 N/m²
- Energy Density (U): \( 6.4 \times 10^{-4} \, \text{J/m}^3 \)