If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased ?

Explanation

To solve the problem of how much the pressure amplitude increases when the sound level in a room is raised from 50 dB to 60 dB, we can follow these steps:

### Step 1: Understand the relationship between sound level and intensity
The sound level in decibels (dB) is given by the formula:
\[ \beta = 10 \log_{10} \left(\frac{I}{I_0}\right) \]
where \(I\) is the intensity of the sound, and \(I_0\) is a reference intensity, typically \(10^{-12} \, \text{W/m}^2\).

### Step 2: Calculate the intensity at 50 dB
Given \(\beta_1 = 50 \, \text{dB}\):
\[ 50 = 10 \log_{10} \left(\frac{I_1}{10^{-12}}\right) \]
Dividing both sides by 10:
\[ 5 = \log_{10} \left(\frac{I_1}{10^{-12}}\right) \]
Taking the antilogarithm:
\[ 10^5 = \frac{I_1}{10^{-12}} \]
Thus, multiplying both sides by \(10^{-12}\):
\[ I_1 = 10^5 \times 10^{-12} = 10^{-7} \, \text{W/m}^2 \]

### Step 3: Calculate the intensity at 60 dB
Now, for \(\beta_2 = 60 \, \text{dB}\):
\[ 60 = 10 \log_{10} \left(\frac{I_2}{10^{-12}}\right) \]
Dividing both sides by 10:
\[ 6 = \log_{10} \left(\frac{I_2}{10^{-12}}\right) \]
Taking the antilogarithm:
\[ 10^6 = \frac{I_2}{10^{-12}} \]
Thus, multiplying both sides by \(10^{-12}\):
\[ I_2 = 10^6 \times 10^{-12} = 10^{-6} \, \text{W/m}^2 \]

### Step 4: Relate intensity to pressure amplitude
The intensity of sound is proportional to the square of the pressure amplitude:
\[ \frac{I_2}{I_1} = \left(\frac{P_2}{P_1}\right)^2 \]
Substituting the values of \(I_1\) and \(I_2\):
\[ \frac{10^{-6}}{10^{-7}} = \left(\frac{P_2}{P_1}\right)^2 \]
Simplifying the left side:
\[ 10 = \left(\frac{P_2}{P_1}\right)^2 \]

### Step 5: Solve for the pressure amplitude ratio
Taking the square root of both sides:
\[ \frac{P_2}{P_1} = \sqrt{10} \approx 3.16 \]

### Conclusion
The pressure amplitude increases by a factor of approximately 3.16 when the sound level is increased from 50 dB to 60 dB.

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