Two sounds in a gas differ in their loudness level by 20 db. What is the ratio of the amplitudes of the pressure oscillations ?

Options

A. A. `10:1`
B. B. `10^4:1`
C. C. `sqrt10:1`
D. D. `100:1`

Answer

To solve the problem of finding the ratio of the amplitudes of pressure oscillations when two sounds differ in their loudness level by 20 dB, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Loudness Level (dB)**:
The loudness level (ΔL) in decibels is given by the formula:
\[ \Delta L = 10 \log_{10} \left( \frac{I_2}{I_1} \right) \]
where \(I_2\) and \(I_1\) are the intensities of the two sounds.

2. **Substituting the Given Value**:
We know from the problem that:
\[ \Delta L = 20 \, \text{dB} \]
Substituting this into the equation gives:
\[ 20 = 10 \log_{10} \left( \frac{I_2}{I_1} \right) \]

3. **Simplifying the Equation**:
Dividing both sides by 10:
\[ 2 = \log_{10} \left( \frac{I_2}{I_1} \right) \]

4. **Exponentiating to Remove the Logarithm**:
To eliminate the logarithm, we exponentiate both sides:
\[ \frac{I_2}{I_1} = 10^2 = 100 \]

5. **Relating Intensity to Pressure Amplitude**:
The intensity \(I\) of a sound wave is proportional to the square of the pressure amplitude \(P\):
\[ I \propto P^2 \]
Therefore, we can write:
\[ \frac{I_2}{I_1} = \frac{P_2^2}{P_1^2} \]

6. **Setting Up the Ratio of Pressure Amplitudes**:
From the intensity ratio we found:
\[ 100 = \frac{P_2^2}{P_1^2} \]
Taking the square root of both sides gives:
\[ \frac{P_2}{P_1} = \sqrt{100} = 10 \]

7. **Final Result**:
Thus, the ratio of the amplitudes of the pressure oscillations is:
\[ P_2 : P_1 = 10 : 1 \]

### Conclusion:
The ratio of the amplitudes of the pressure oscillations is \(10 : 1\).

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