If you measure the sound pressure level from a certain distance, you can recalculate them to another distance. But what is, if a noise is generated directly on your membrane (microphone membrane or eardrum for example). How is the sound pressure calculated in this situation? Because the distance from the "sound level meter membrane" is zero in this case.


2 Answers 2


You can't. For the calculation of pressure anomalies you need a volume of a medium in which the sound propagates. Apart from that you need a non-zero distance to apply the inverse-distance-squared law.

If you generate the sound on the microphone, you don't have to calculate the intensity. You can simply read the oscillations from the sound generator...and buy a new microphone.

  • $\begingroup$ But how loud would it be approximately in comparison to a very very short distance? Because if I go for the distance law it would be infinite :) $\endgroup$
    – J. Scott
    Commented Feb 28, 2018 at 16:34
  • 1
    $\begingroup$ @J.Scott The point of the answer is that you can't apply the law with no distance. It doesn't hold under that condition. $\endgroup$
    – JMac
    Commented Feb 28, 2018 at 19:35
  • $\begingroup$ Ok thank you all. But I have just one more question: What is, if I measure a sound from a normal distance (in the free field). Can I than calculate it back to the near field (0.025m)? Does the assumption 1 from @D. Betchkal only refers to a measurement in the near field or also for calculations to the near field? $\endgroup$
    – J. Scott
    Commented Mar 2, 2018 at 18:39
  • $\begingroup$ To clarify the inverse distance law... You can use it only in the far-field (pressure and particle velocity are in phase) otherwise (near field) you could get different results for the same distance from the source for different coordinates of measurement (there can be cancellations due to different parts of the source moving in different phases). For a simple case, you can consider an ideal dipole source and search for the "acoustical short-circuit" term. $\endgroup$
    – ZaellixA
    Commented Feb 1, 2020 at 14:13

I'd like to add to Aziraphale's answer by describing the assumptions that must hold for the most basic spreading loss equations to be effective in predicting Sound Pressure Level (SPL).

Because SPL is a field quantity, the assumptions are spatial, but they cannot be crisply defined for the general case. They'll depend on the physical system in question.

The following diagrams shows the two assumptions that must hold to predict SPL with a simple spreading loss equation:

enter image description here

Assumption 1:

(To answer your question) the plot shows that if you get really close to a source the Far Field assumption starts to break down. The simple way to say this is the relationship of SPL to distance is complex.

The more technical answer is that the Far Field assumption breaks down when acoustic pressure $p$ and acoustic particle velocity $v$ are no longer in phase with each other.

In practice the near field boundary is generally "2 - 3x the wavelength of interest". If you start to think about it, if you make measurements of sound pressure level close enough to a source such that you're less than two wavelengths for audible sounds you'd be somewhere between 8 meters (80 Hz) and 0.04 m (17150 Hz) away from the source. It should be apparent that the physical form of the source will begin to have important influences on the sound field when you're that close.

More simply put, it's harder to assume the sound field is perfectly spherical (or plane).

I want to point out that it is useful to measure sound intensity $I$ in the near field, and in fact, it's one of the main ways scientists estimate the sound power of a source (pg 14).

Assumption 2:

The plot also shows that if you get really far from a source the Free Field assumption starts to break down. In other words we can no longer assume

$$p_{direct} \gg p_{reflected} $$

Such that reflections are no longer a negligible part of the overall acoustic field. You can see that depending on the physical form of the reverberant space you can have a variety of rates of fall off, with the most extreme example being something like a Reverberation Room, where the sound field becomes essentially constant after a certain distance away from the source.


if you're too close or too far from your source the basic spreading loss equations no longer apply
  • $\begingroup$ Thank you very much for your answer. So, i can assume, that the distance law does not fit for very short (<0,04m) and very long (>8m) distances? One example: If we have a broadband noise, that is 120 dB SPL loud from a distance of 20 cm. How loud would it be approximately, if the same noise would be generated directly on the microphone membrane, which measures the noise? Can we estimate the sound intensity from the given sound pressure level? $\endgroup$
    – J. Scott
    Commented Feb 28, 2018 at 21:25
  • $\begingroup$ D. Betchkal, can I write you a PM, please. I need some help in acoustics. I would be very grateful for that. $\endgroup$
    – J. Scott
    Commented Feb 28, 2018 at 21:59
  • $\begingroup$ Sound intensity is independent of the distance, right? But why is it the same as the sound pressure level in 0,2821m, according to this link: sengpielaudio.com/calculator-soundpower.htm $\endgroup$
    – J. Scott
    Commented Feb 28, 2018 at 22:58
  • 1
    $\begingroup$ Nope, sound Intensity is not independent of distance from source: physics.stackexchange.com/questions/387138/… Intensity is the energy flow per unit surface area, and so as the wavefront spreads out with distance, the total amount of energy stays the same but the area increases. Sound power level is independent of distance. $\endgroup$ Commented Feb 28, 2018 at 23:24

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