For a plane harmonic sound wave of frequency 1000 Hz in air, the pressure amplitude is `"1 N/m"^2?` What is the displacement amplitude ? Atmospheric pressure is `1.01 xx 10^(5)" N/m"^2, gamma = 1.4.` Velocity of sound in air is `340 ms^(-1)?.`

Answer

To find the displacement amplitude of a plane harmonic sound wave given the pressure amplitude, we can follow these steps:

### Step 1: Identify the given values
- Frequency (f) = 1000 Hz
- Pressure amplitude (ΔP) = 1 N/m²
- Atmospheric pressure (P₀) = 1.01 x 10⁵ N/m²
- Velocity of sound in air (v) = 340 m/s
- Gamma (γ) = 1.4

### Step 2: Calculate the effective pressure (E)
The effective pressure (E) can be calculated using the formula:
\[ E = \gamma \times P₀ \]
Substituting the values:
\[ E = 1.4 \times 1.01 \times 10^5 \]
\[ E = 1.414 \times 10^5 \, \text{N/m}² \]

### Step 3: Calculate the wavelength (λ)
The wavelength (λ) can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
Substituting the values:
\[ \lambda = \frac{340 \, \text{m/s}}{1000 \, \text{Hz}} \]
\[ \lambda = 0.34 \, \text{m} \]

### Step 4: Calculate the displacement amplitude (A)
The displacement amplitude (A) can be calculated using the formula:
\[ A = \frac{\Delta P \cdot \lambda}{2 \pi E} \]
Substituting the values:
\[ A = \frac{1 \, \text{N/m}² \cdot 0.34 \, \text{m}}{2 \pi \cdot (1.414 \times 10^5 \, \text{N/m}²)} \]

### Step 5: Simplify the calculation
Calculating the denominator:
\[ 2 \pi \cdot (1.414 \times 10^5) \approx 8.88 \times 10^5 \]
Now substituting back:
\[ A = \frac{0.34}{8.88 \times 10^5} \]
\[ A \approx 3.83 \times 10^{-7} \, \text{m} \]

### Final Result
The displacement amplitude is approximately:
\[ A \approx 3.8 \times 10^{-7} \, \text{m} \]

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