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In a completely open space, we position one speaker.

The speaker emits a sound at X dB.

How can I calculate the sound's volume at a specific distance?

Also, how can I calculate the attenuation of a specific frequency range for a specific distance? my guess being that all frequencies will not decay the same way, but I could be wrong.

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    $\begingroup$ Use the inverse square law: hyperphysics.phy-astr.gsu.edu/hbase/acoustic/isprob2.html $\endgroup$
    – Ernie
    Commented Dec 14, 2015 at 3:15
  • $\begingroup$ The inverse square law seems to be a bit problematic since The output of the speaker is quite directional. Its attenuation, which has directional dependence, can in principle be calculated using the relevant wave equation but that can be complicated. Frequency dependence enters through the frequency dependence of the sound velocity. $\endgroup$
    – Urgje
    Commented Dec 14, 2015 at 9:43

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There are (at least) two effects to consider:

First - a point source of sound in free space (air on all sides) and that emits power isotropically (in all directions) will see power diminish with distance (since the "sphere" of sound gets bigger, the sound at any one point on that sphere must have less energy). This results in a decrease of 6 dB every time you double the distance. Since most sound sources are neither isotropic, nor point sources, you usually can start using this approximation once you are "far" compared to the size of the source (loudspeaker, for example). So if you have a 20 cm diameter speaker, you can say that if the sound level is 66 dB at 4 m, it will be 60 dB at 8 m and 54 dB at 16 m. But you can't say that it will be 84 dB at 50 cm - you are getting "close compared to the size of the source".

Second - attenuation of individual frequencies due to attenuation. This is much more complex; I described some of the factors in this earlier answer, which shows that the attenuation of sound in air is a function of temperature, humidity, and frequency - among other factors. As a rule of thumb, higher frequencies will be attenuated more strongly. And that effect is linear with distance: if you have a 3 dB attenuation per 10 m, you will have 30 dB over 100 m. In that way it is different than the inverse square relationship, which becomes less significant as you move further away (since you have to move much further to double the distance again).

Another way to think about that: as long as the wave front looks like a sphere, the inverse square law is very important; but once the wave front starts to look like a plane wave, linear attenuation becomes the driving factor.

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  • $\begingroup$ About the first point: what would define 'far' or 'close' compared to the size of the source? $\endgroup$
    – Thomas
    Commented Dec 17, 2015 at 16:15
  • $\begingroup$ Rule of thumb: Ten times the diameter of the source is "far"; less that three times the diameter is "close". In between - well... $\endgroup$
    – Floris
    Commented Dec 17, 2015 at 17:28
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    $\begingroup$ ok; this gave me great information to continue my work, I'll mark this post as the answer. $\endgroup$
    – Thomas
    Commented Dec 17, 2015 at 20:44
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If it is in open space (as in in a vacuum) sound wouldn't travel. If you mean in an open space here on earth, the sound volume would depend on temp of the air, humidity, pressure, etc.

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