Question

A rock group is playing in a bar. Sound emerging from the door spreads uniformly in all directions. The intensity level of the music is 92 dB at a distance of 3.52 m from the door. At what distance is the music just barely audible to a person with a normal threshold of hearing

Answer 1

The decibel scale is given by:

$dB=10\log _{10}(\frac{I}{I_0})$

where I is the sound intensity and I0 is the threshold intensity; plugging the value given for the music level we have that:

$\begin{gathered} 92=10\log _{10}(\frac{I}{1\times10^{-12}}) \\ \log _{10}(\frac{I}{1\times10^{-12}})=\frac{92}{10} \\ \frac{I}{1\times10^{-12}}=10^{\frac{92}{10}} \\ I=1\times10^{-12}\cdot10^{\frac{92}{10}} \\ I=1.58\times10^{-3} \end{gathered}$

We know that sound intensity is inversely proportional to the square of the distance of the source, then we have:

$\frac{I_1}{I_2}=\frac{r^2_2}{r^2_1}$

Plugging the values we found:

$\begin{gathered} \frac{1.58\times10^{-3}}{1\times10^{-12}}=\frac{r^2_2}{(3.52)^2} \\ r^2_2=(3.52)^2(\frac{1.58\times10^{-3}}{1\times10^{-12}}) \\ r_2=\sqrt[]{(3.52)^2(\frac{1.58\times10^{-3}}{1\times10^{-12}})} \\ r_2=139917 \end{gathered}$

Therefore, the distance is 139917 m. (This is a large distance since sound is not being absorbed)