A BIRD SINGS IN A MEADOW
Consider an idealized bird (treated as a point source) that emits constant sound power, with intensity obeying the inverse-square law (Fig. 16.11). If you move twice the distance from the bird, by how many decibels does the sound intensity level drop?
IDENTIFY and SET UP:
The decibel scale is logarithmic, so the difference between two sound intensity levels (the target variable) corresponds to the ratio of the corresponding intensities, which is
determined by the inverse-square law. We label the two points P_1 and P_1 (Fig. 16.11). We use Eq. (16.15), the definition of sound intensity level, at each point. We use Eq. (15.26), the inverses-quare law, to relate the intensities at the two points.
\beta=(10 \mathrm{~dB}) \log \frac{I}{I_{0}} (16.15)
\frac{I_{1}}{I_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}} (15.26)
EXECUTE:
The difference \beta_{2}-\beta_{1} between any two sound intensity levels is related to the corresponding intensities by
\begin{aligned}\beta_{2}-\beta_{1} &=(10 \mathrm{~dB})\left(\log \frac{I_{2}}{I_{0}}-\log\frac{I_{1}}{I_{0}}\right) \\&=(10 \mathrm{~dB})\left[\left(\log I_{2}-\log I_{0}\right)-\left(\log I_{1}-\log I_{0}\right)\right] \\&=(10 \mathrm{~dB}) \log \frac{I_{2}}{I_{1}}\end{aligned}For this inverse-square-law source, Eq. (15.26) yields I_{2} / I_{1}=r_{1}^{2} / r_{2}^{2}=\frac{1}{4}, So
\beta_{2}-\beta_{1}=(10 \mathrm{~dB}) \log \frac{I_{2}}{I_{1}}=(10 \mathrm{~dB}) \log \frac{1}{4}=-6.0 \mathrm{~dB}
EVALUATE: Our result is negative, which tells us (correctly) that the sound intensity level is less at P_2 than at P_1. The 6-dB difference doesn’t depend on the sound intensity level at P_1; any doubling of the distance from an inverse-square-law source reduces the sound intensity level by 6 dB.
Note that the perceived loudness of a sound is not directly proportional to its intensity. For example, most people interpret an increase of 8 dB to 10 dB in sound intensity level (corresponding to increasing intensity by a factor of 6 to 10) as a doubling of
loudness.