Question

A woman stands a distance d from a loud motor that emits sound uniformly in all directions. The sound intensity at her position is an uncomfortable 3.9×10-3 W/m2. At a distance 3.8 times as far from the motor, what are (a) the sound intensity and (b) the sound intensity level relative to the threshold of hearing?

Answer 1

From the general equations of sound we know that the Intensity is directly proportional to the square of the distance. Therefore the relationship,

$I \propto \frac{1}{r^2}$

It will help us find the change between the two points. If the relationship is maintained, the intensity of point one versus point two will be given by the square of the distance of the other two points, therefore

$\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}$

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

From the statement it is given that the distance is 3.8, therefore

$r_2 = 3.8r_1$

and

$I_1 = 3.9*10^{-3}W/m^2$

PART A) The intensity given under the previous equation found is

$I_2 = 3.9*10^{-3} (\frac{1}{3.8})^2$

$I_2 = 0.00027W/m^2$

Therefore the sound intensity at point 2 is 0.00027W/m^2

PART B) For the sound level in decibels we will apply the logarithmic equation that relates the intensities, therefore

$\beta = 10log_{10} (\frac{I}{I_0})$

Where $I_0$ is the threshold intensity and is given by

$I_0 = 10^{-12} W/m^2$

Then,

$\beta = 10log_{10} (\frac{0.00027}{10^{-12}})$

$\beta = 84.31dB$

Therefore the sound intensity level relative to the threshold of hearing is 84.31dB