Question

In general, if a sound has intensity of β dB at 1 m from the source, at what distance from the source would the decibel level decrease to 0 dB? Since the limit of hearing is 1 dB this would mean you could no longer hear it. Express the answer in terms of β. Possibly relevant information: The sound of normal breathing is not very loud, with an intensity of about 11 dB at a distance of 1 m away from the face of the breather. Again: EXPRESS THE ANSWER IN TERMS OF β!

Answer 1

Answer: The required distance is given by

$r_2=1\text{ m}\cdot 10^\frac{\beta}{20}.$

Explanation: The sound intensity in dB is given by the formula

$\beta \text{ dB}=10\log\frac{I}{I_0}$

where $I_0$ is the hearing threshold in absolute units and $I$ is the absolute intensity of the sound which depends on the distance. In general, for two distances $r_1$ and $r_2$ we have that

$\frac{I(r_1)}{I(r_2)}=\frac{r_2^2}{r_1^2}=\left(\frac{r_2}{r_1}\right)^2.$

Now let us take $r_1=1\text{ m}$ and let $r_2$ be the required distance. We have

$\beta \text{ dB}=10\log\frac{I(r_1)}{I_0},\quad 0\text{ db}=10\log\frac{I(r_2)}{I_0}.$

Exponentiating these equations we obtain

$10^{\frac{\beta}{10}}=\frac{I(r_1)}{I_0},\quad 10^0=1=\frac{I(r_2)}{I_0}.$

Dividing them

$\frac{I(r_1)}{I(r_2)}=10^\frac{\beta}{10}.$

Using the previously stated identity

$\frac{r_2^2}{r_1^2}=10^\frac{\beta}{10}\Rightarrow r_2=r_110^\frac{\beta}{20}=1\text{ m}\cdot 10^\frac{\beta}{10}$

Now if we use the given example where $\beta=11$ we have

$r_2=1\text{ m}\cdot 10^{\frac{11}{20}}=3.55\text{ m}.$