If the intensity of sound is doubled, by how many decibels does the sound level increase ?

Explanation

To solve the problem of how many decibels the sound level increases when the intensity of sound is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Sound Level in Decibels:**
The sound level (β) in decibels is given by the formula:
\[ \beta = 10 \log \left(\frac{I}{I_0}\right) \]
where \(I\) is the intensity of the sound and \(I_0\) is the reference intensity.

2. **Calculate the Initial Sound Level (β1):**
Let the initial intensity be \(I\). The initial sound level (β1) can be calculated as:
\[ \beta_1 = 10 \log \left(\frac{I}{I_0}\right) \]

3. **Calculate the New Sound Level (β2) When Intensity is Doubled:**
When the intensity is doubled, the new intensity becomes \(2I\). The new sound level (β2) is:
\[ \beta_2 = 10 \log \left(\frac{2I}{I_0}\right) \]

4. **Find the Change in Sound Level (Δβ):**
The change in sound level (Δβ) is given by:
\[ \Delta \beta = \beta_2 - \beta_1 \]
Substituting the expressions for β2 and β1:
\[ \Delta \beta = 10 \log \left(\frac{2I}{I_0}\right) - 10 \log \left(\frac{I}{I_0}\right) \]

5. **Use Logarithmic Properties:**
Using the property of logarithms that states \(\log a - \log b = \log \left(\frac{a}{b}\right)\), we can simplify:
\[ \Delta \beta = 10 \left( \log \left(\frac{2I}{I_0}\right) - \log \left(\frac{I}{I_0}\right) \right) = 10 \log \left(\frac{2I/I_0}{I/I_0}\right) \]
This simplifies to:
\[ \Delta \beta = 10 \log \left(2\right) \]

6. **Calculate the Value of log(2):**
The value of \(\log(2)\) is approximately \(0.3010\). Thus:
\[ \Delta \beta = 10 \times 0.3010 = 3.010 \]

7. **Final Result:**
Therefore, the increase in sound level when the intensity is doubled is approximately:
\[ \Delta \beta \approx 3 \text{ decibels} \]

### Summary:
When the intensity of sound is doubled, the sound level increases by approximately 3 decibels.