The inverse square law and the sound intensity - sengpielaudio
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Sound intensity I and the inverse square law 1/r2
How does the sound intensity decrease with distance from the sound source?
How does the sound intensity level decrease with doubling the distance from the source?
Sound pressure is not sound intensity. p2I for progressive plane waves.

Compare also the distance law 1/r, when
using sound field values, e.g. sound pressure

In the real world, the inverse square law (squared distance law) I ∝ 1/r2 is always an idealization because it assumes exactly equal sound intensity I as sound energy propagation in all directions. If there are reflective surfaces in the sound field, then reflected sounds will add to the directed sound and you will get more sound intensity at a field location than the inverse square law predicts. If there are barriers between the source and the point of measurement, you may get less than the square law predicts. Nevertheless, the inverse square law is the logical first estimate of the sound intensity you would get at a distant point in a reasonably open area. The reference sound intensity level SIL = 0 dB is the sound intensity of I0 = 1 µW/m2 = 1 × 10-12 W/m2.


If you measure at distance
r1   = m = ft
a sound intensity level (SIL1)
Lp1 = dB,
then at distance
r2   = m = ft
the inverse square law 1/r2 predicts
a sound intensity level (SIL2)

Lp2 = dB.
You can explore numerically to confirm that doubling the distance drops the intensity I to a quarter (0.25) of the initial value and the intensity level by about 6 dB and that 10 times the distance drops the intensity I to a hundredth (0.01) of the inititial value and the intensity level by about 20 dB.

Enter a value in the left or right box, then press the TAB bar or make
a mouse click at an empty space at the side, to get the solution.
The calculator works in both directions of the
sign.
Sound intensity level LI:
dB-SIL
 ↔  Sound intensity  I:
W/m2
Start   Start
Standard reference sound intensity I0 = 1 pW/m2 or 10-12 W/m2 (0 dB)
Inverse square law 1/r2

Inverse Square Law

Law for Sound Energy Quantities
Distance ratio Sound Intensity I ∝ 1/ r²
1   1/1² = 1/1 = 1.0000
2   1/2² = 1/4 = 0.2500
3   1/3² = 1/9 = 0.1111
4 1/4² = 1/16 = 0.0625
5 1/5² = 1/25 = 0.0400
6 1/6² = 1/36 = 0.0278
7 1/7² = 1/49 = 0.0204
8 1/8² = 1/ 64 = 0.0156
9 1/9² = 1/81 = 0.0123
10 1/10² =1/100 = 0.0100

Calculating sound intensity with the inverse square law

I1 / I2 = r22 / r12

I2 = I1 (r12 / r22)

Where:
I1  =  sound intensity 1 at r1
I2  =  sound intensity 2 at r2
r1  =  distance 1 from source
r2  =  distance 2 from source

A doubling of distance from the sound source in the direct field will reduce the "sound level"
by 6 dB, no matter whether that are sound intensity levels or sound pressure levels! This will
reduce the sound intensity I (energy quantity) to 1/2² = 1/4 (25 %) and the sound pressure p
(field quantity) to 1/2 (50 %) of the the initial value.

The inverse square law 1/r2 shows the distance performance of energy quantities and
the inverse distance law 1/r shows the distance performance of field quantities.
Energy quantities are propotional to squared field quantities; I p2

Sound Energy Quantities
Sound intensity, sound energy density, sound power, electric power.
 
Inverse Square Law 1/r²
Sound Field Quantities   :-)
Sound pressure, sound or particle velocity, particle displacement or particle ampltide, voltage, current, electric resistance.
Inverse Distance Law 1/r

Conversions and Calculations - Sound Quantities and their Levels


Frequently used wrong statements in the context of sound sizes

Wrong expressions Correct version
Sound intensity decreases inversely as the
distance increases with 1/r
from the sound source.
Sound intensity decreases inversely as the square
of the distance increases with 1/r2
from the sound
source.
Sound intensity level decreases inversely as the
square of the distance increases with 1/r2 from the
sound source. 3 dB per doubling.
Sound intensity level decreases by 6 dB per doubling
of distance from the source to 1/4 (25 %) of the sound
intensity initial value.
Sound pressure decreases inversely as the square
of the distance increases with 1/r2
from the
sound source.
Sound pressure decreases inversely as the
distance increases with 1/r
from the sound source.
Sound pressure level decreases inversely as the
square of the distance increases with 1/r2
from the
sound source. 3 dB per doubling.
Sound pressure level decreases by 6 dB per doubling
of distance from the source to 1/2 (50 %) of the sound
pressure initial value.

Neither the sound power, nor the sound power level decreases in
doubling the distance up to a value or up to any dB. Why is this so?

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