An isotropic point source 'S' of sound emits constant power. Two points A and B are situated at distance x and 2x from S. the difference between the loudness of points A and B is about (log 2=0.3)

Options

A. 3 dB
B. 2 dB
C. 6 dB
D. 12 dB

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding Loudness and Intensity
Loudness (β) is related to the intensity (I) of sound by the formula:
\[ \beta = 10 \log \left( \frac{I}{I_0} \right) \]
where \( I_0 \) is the reference intensity, typically \( 10^{-12} \, \text{W/m}^2 \).

### Step 2: Define Loudness at Points A and B
Let’s denote the loudness at point A as \( \beta_A \) and at point B as \( \beta_B \):
\[ \beta_A = 10 \log \left( \frac{I_A}{I_0} \right) \]
\[ \beta_B = 10 \log \left( \frac{I_B}{I_0} \right) \]

### Step 3: Calculate the Difference in Loudness
The difference in loudness between points A and B can be expressed as:
\[ \beta_A - \beta_B = 10 \log \left( \frac{I_A}{I_B} \right) \]

### Step 4: Relate Intensity to Distance
For a point source, the intensity is inversely proportional to the square of the distance from the source:
\[ I \propto \frac{P}{4\pi r^2} \]
Thus, we can say:
\[ I_A \propto \frac{P}{x^2} \quad \text{and} \quad I_B \propto \frac{P}{(2x)^2} = \frac{P}{4x^2} \]

### Step 5: Express Intensities
From the above relationships, we can express the intensities:
\[ I_A = \frac{P}{x^2} \quad \text{and} \quad I_B = \frac{P}{4x^2} \]

### Step 6: Substitute Intensities into the Loudness Difference Formula
Now substituting \( I_A \) and \( I_B \) into the loudness difference:
\[ \beta_A - \beta_B = 10 \log \left( \frac{I_A}{I_B} \right) = 10 \log \left( \frac{\frac{P}{x^2}}{\frac{P}{4x^2}} \right) \]

### Step 7: Simplify the Expression
This simplifies to:
\[ \beta_A - \beta_B = 10 \log \left( \frac{4}{1} \right) = 10 \log(4) \]

### Step 8: Use Logarithmic Properties
Using the property of logarithms, we can rewrite \( \log(4) \) as:
\[ \log(4) = \log(2^2) = 2 \log(2) \]
Thus:
\[ \beta_A - \beta_B = 10 \cdot 2 \log(2) = 20 \log(2) \]

### Step 9: Substitute the Value of \( \log(2) \)
Given \( \log(2) = 0.3 \):
\[ \beta_A - \beta_B = 20 \cdot 0.3 = 6 \, \text{dB} \]

### Conclusion
The difference between the loudness of points A and B is:
\[ \beta_A - \beta_B = 6 \, \text{dB} \]