Due to point isotropic sound source, the intensity at a point is observed as 40 dB. The density of air is `rho=((15)/(11))(kg)/(m^3)` and velocity of sound in air is `330(m)/(s)`. Based on this information answer the following questions.
Q. The ratio of displacement amplitude of wave at observation point to wavelength of sound waves is

Options

A. `3.22xx10^-6`
B. `3.22xx10^-12`
C. `3.22xx10^-9`
D. `1.07xx10^-10`

Explanation

To solve the problem, we need to find the ratio of the displacement amplitude of the wave at the observation point to the wavelength of the sound waves. We will follow these steps:

### Step 1: Calculate the Intensity (I)
Given that the sound intensity level (β) is 40 dB, we can use the formula for sound intensity level:

\[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]

Where \( I_0 = 10^{-12} \, \text{W/m}^2 \) is the reference intensity. Rearranging the formula to find \( I \):

\[ 40 = 10 \log_{10} \left( \frac{I}{10^{-12}} \right) \]

Dividing both sides by 10:

\[ 4 = \log_{10} \left( \frac{I}{10^{-12}} \right) \]

Taking the antilogarithm:

\[ \frac{I}{10^{-12}} = 10^4 \]

Thus,

\[ I = 10^4 \times 10^{-12} = 10^{-8} \, \text{W/m}^2 \]

### Step 2: Calculate the Pressure Amplitude (ΔP)
The relationship between intensity and pressure amplitude is given by:

\[ I = \frac{(\Delta P)^2}{2 \rho v} \]

Rearranging for ΔP:

\[ \Delta P = \sqrt{2 I \rho v} \]

Substituting the known values:
- \( I = 10^{-8} \, \text{W/m}^2 \)
- \( \rho = \frac{15}{11} \, \text{kg/m}^3 \)
- \( v = 330 \, \text{m/s} \)

Calculating ΔP:

\[ \Delta P = \sqrt{2 \times 10^{-8} \times \frac{15}{11} \times 330} \]

Calculating the values inside the square root:

\[ \Delta P = \sqrt{2 \times 10^{-8} \times 1.3636 \times 330} \]

Calculating:

\[ \Delta P \approx \sqrt{8.99 \times 10^{-6}} \approx 3 \times 10^{-3} \, \text{N/m}^2 \]

### Step 3: Calculate the Bulk Modulus (B)
The speed of sound is related to the bulk modulus and density by:

\[ v = \sqrt{\frac{B}{\rho}} \]

Rearranging gives:

\[ B = v^2 \rho \]

Substituting the known values:

\[ B = (330)^2 \times \frac{15}{11} \]

Calculating:

\[ B = 108900 \times 1.3636 \approx 148000 \, \text{Pa} \]

### Step 4: Calculate the Ratio of Displacement Amplitude to Wavelength (A/λ)
Using the relationship:

\[ \Delta P = B \cdot A \cdot \frac{2\pi}{\lambda} \]

Rearranging gives:

\[ \frac{A}{\lambda} = \frac{\Delta P}{B \cdot 2\pi} \]

Substituting the values:

\[ \frac{A}{\lambda} = \frac{3 \times 10^{-3}}{148000 \cdot 2\pi} \]

Calculating:

\[ \frac{A}{\lambda} = \frac{3 \times 10^{-3}}{148000 \cdot 6.2832} \approx \frac{3 \times 10^{-3}}{930000} \approx 3.22 \times 10^{-9} \]

### Final Answer
Thus, the ratio of displacement amplitude of the wave at the observation point to the wavelength of sound waves is approximately:

\[ \frac{A}{\lambda} \approx 3.22 \times 10^{-9} \]

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