Sound Intensity & Distance Relationship

[karush]

4. Sound intensity is inversely proportional to the square of the distance from the source
The further from the source you are, the less intense the sound.
Suppose the sound intensity is 20 watts per square meter. (W/m^2) at 8 meters.
What is the sound intensity at 4 meters?

$y=\frac{k}{x}$ So. $s=\frac{k}{{d}^{2}}$

$8=\frac{k}{20}$ then $k=160$

$s=\frac{160}{{4}^{2}}=10 W/m^2$

I wasn't sure how this was all pluged in?

 

[MarkFL]

Let the intensity $I$ be measured in units of \(\displaystyle \frac{\text{W}}{\text{m}^2}\), where we are given the inverse-square law of radiation (which works with any kind of energy propagated in 3-space, at least for classical mechanics):

\(\displaystyle I(r)=\frac{k}{r^2}\)

where $k$ is a constant of proportionality, and $r$ is the distance from the source. We are given:

\(\displaystyle I(8)=20=\frac{k}{8^2}\implies k=1280\)

And so we have:

\(\displaystyle I(r)=\frac{1280}{r^2}\)

Thus:

\(\displaystyle I(4)=\frac{1280}{4^2}=80\)

We should expect that if we divide our distance from the source by some factor $0<a$, then the intensity will change by a factor of $a^2$, since:

\(\displaystyle \frac{\dfrac{k}{\left(\dfrac{r}{a}\right)^2}}{\dfrac{k}{r^2}}=a^2\)

 

[karush]

I should have seen that..
Mahalo