If the distance to a sound source is halved, how will the sound intensity level change?
a) increase by a factor of 2
b) depends on the actual distance
c) increase by a factor of 4.
d) increase by 6 dB
e) increase by 3 dB
Loosely speaking; sound intensity of a source is measured in terms of theOriginally Posted by Candy
energy flow across unit surface area perpendicular to the wave front.
Lets assume we are in the spherical spreading regime, then the surface
area of a sphere of radius $\displaystyle R$ increases as $\displaystyle R^2$, so the energy density
crossing a spherical shell of radius $\displaystyle R$ goes as $\displaystyle 1/R^2$ (energy/power is
the same crossing shells of all radii, but the area it is passing across
increases as $\displaystyle R^2$).
So halving the distance from the source quadruples the sound intensity.
However you are asked about the sound intensity level, which is the
sound intensity expressed in dB. For intensity levels the definition is:
$\displaystyle L_I=10\ \log_{10}\left[ \frac{I}{I_0} \right]$,
where $\displaystyle I_0$ is the reference intensity level (which we don't
need to know but is often (and sensibly) $\displaystyle 1 \mbox{ w/m^2 }$.
Now at this point we could do some sums, but we should know that
a factor of four increase in intensity is equivalent to $\displaystyle 6 \mbox{dB}$ increase
in intensity level, which is the answer.
RonL