The pressure variation in a sound wave in air is given by
`Deltap = 12 sin (8.18X - 2700 t + pi//4) N//m^(2)`
find the displacement amplitude. Density of air `= 1.29 kg//m^(3)`.

Explanation

To find the displacement amplitude from the given pressure variation of a sound wave, we will follow these steps:

### Step 1: Identify the parameters from the pressure variation equation
The pressure variation in the sound wave is given by:
\[ \Delta P = 12 \sin(8.18X - 2700t + \frac{\pi}{4}) \, \text{N/m}^2 \]
From this equation, we can identify:
- The pressure amplitude, \(\Delta P_m = 12 \, \text{N/m}^2\)
- The angular frequency, \(\omega = 2700 \, \text{rad/s}\)
- The wave number, \(k = 8.18 \, \text{rad/m}\)

### Step 2: Calculate the speed of sound using the wave number and frequency
The speed of sound \(v\) can be calculated using the relationship:
\[ v = f \lambda \]
where \(f\) is the frequency and \(\lambda\) is the wavelength. We can find \(f\) from \(\omega\):
\[ f = \frac{\omega}{2\pi} = \frac{2700}{2\pi} \approx 429.45 \, \text{Hz} \]
Now, we can find the wavelength \(\lambda\) using the wave number \(k\):
\[ k = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{k} = \frac{2\pi}{8.18} \approx 0.769 \, \text{m} \]
Now, we can calculate the speed of sound:
\[ v = f \lambda = 429.45 \times 0.769 \approx 330 \, \text{m/s} \]

### Step 3: Use the relationship between pressure amplitude and displacement amplitude
The relationship between the pressure amplitude \(\Delta P_m\) and the displacement amplitude \(A\) is given by:
\[ \Delta P_m = \rho v \omega A \]
where \(\rho\) is the density of air (given as \(1.29 \, \text{kg/m}^3\)).

### Step 4: Rearrange the equation to solve for displacement amplitude \(A\)
Rearranging the equation gives:
\[ A = \frac{\Delta P_m}{\rho v \omega} \]

### Step 5: Substitute the known values into the equation
Now substituting the known values:
\[ A = \frac{12}{1.29 \times 330 \times 2700} \]

### Step 6: Calculate the displacement amplitude
Calculating the above expression:
\[ A = \frac{12}{1.29 \times 330 \times 2700} \approx 1.04 \times 10^{-5} \, \text{m} \]

### Final Answer
The displacement amplitude \(A\) is approximately:
\[ A \approx 1.04 \times 10^{-5} \, \text{m} \]

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