To find out by what factor the intensity changes for a given increase in sound level, we need to use the following formula that relates intensity to decibels (dB):
\[ \beta (dB) = 10 \times \log_{10} \left(\frac{I}{I_0}\right) \]
where:
- \(\beta\) represents the increase or decrease in sound intensity level in decibels (dB).
- \(I\) represents the new intensity of the sound.
- \(I_0\) is the reference intensity (usually the threshold of hearing).
For an increase of 3.6 dB in sound, we want to find the factor by which the intensity, \(I\), is greater than the reference intensity, \(I_0\).
Rearranging the decibel formula to solve for the intensity ratio (change factor) \(I / I_0\), we have:
\[ 3.6 = 10 \times \log_{10} \left(\frac{I}{I_0}\right) \]
Divide both sides by 10:
\[ \frac{3.6}{10} = \log_{10} \left(\frac{I}{I_0}\right) \]
Take the antilogarithm (inverse of logarithm) of both sides to solve for \(I / I_0\):
\[ 10^{\frac{3.6}{10}} = 10^{\log_{10} \left(\frac{I}{I_0}\right)} \]
\[ 10^{\frac{3.6}{10}} = \frac{I}{I_0} \]
Using the properties of logarithms, the right side simply becomes \(\frac{I}{I_0}\) because taking the logarithm and then the antilogarithm of a number gives you the number itself.
Now we compute \(10^{\frac{3.6}{10}}\):
\[ 10^{\frac{3.6}{10}} = 10^{0.36} \]
To find the intensity change factor, we calculate \(10^{0.36}\), which equals approximately 2.290867652767773. Therefore, for a 3.6 dB increase, the intensity of the sound is about 2.290867652767773 times greater than the original intensity.
