In acoustics, sound intensity and sound level are related logarithmically. The table shows several values.
<table> <tr> <td>Sound intensity \(I\) (in \(\text{W}/\text{m}^2\))</td> <td>\(10^{-12}\)</td> <td>\(10^{-10}\)</td> <td>\(10^{-8}\)</td> <td>\(10^{-6}\)</td> </tr> <tr> <td>Sound level \(L\) (in \(\text{dB}\))</td> <td>\(0\)</td> <td>\(20\)</td> <td>\(40\)</td> <td>\(60\)</td> </tr> </table>
In the formula below, \(I\) means the numerical intensity reading in \(\text{W}/\text{m}^2\), and \(L\) is the numerical sound-level reading in decibels.
a) By how many decibels does \(L\) increase when \(I\) is multiplied by \(10\)?
b) The relationship can be modeled by \(L=a\log(I)+b\). Find \(a\) and \(b\), and verify the model using another pair from the table.
c) Find the inverse formula for intensity in \(\text{W}/\text{m}^2\) as a function of \(L\). What intensity corresponds to \(50\,\text{dB}\)?
Hints
- What happens to the sound level when intensity rises by two powers of ten?
- How does the common logarithm change under multiplication by ten?
- Use two table columns to find the coefficients.
- To invert, what operation undoes a common logarithm?
Solution
1. The table shows that multiplying intensity by \(100\) increases the level by \(20\,\text{dB}\), so a factor of \(10\) increases it by \(10\,\text{dB}\).
2. For \(I=10^{-12}\), \(0=-12a+b\), so \(b=12a\).
3. For \(I=10^{-10}\), \(20=-10a+12a\), so \(a=10\), \(b=120\), and \(L=10\log(I)+120\).
4. Check \(I=10^{-6}\): \(L=10(-6)+120=60\), matching the table.
5. Solve \(L-120=10\log(I)\) for the numerical intensity: \(I=10^{(L-120)/10}\). Thus, the physical intensity is \(10^{(L-120)/10}\,\text{W}/\text{m}^2\).
6. At \(L=50\), the intensity is \(10^{-7}\,\text{W}/\text{m}^2\).
Answer
a) \(10\,\text{dB}\)
b) \(a=10\), \(b=120\), so \(L=10\log(I)+120\); the \(10^{-6}\) reading gives \(60\,\text{dB}\).
c) Physical intensity \(=10^{(L-120)/10}\,\text{W}/\text{m}^2\); at \(50\,\text{dB}\), intensity \(=10^{-7}\,\text{W}/\text{m}^2\).