Based on this answer: Relationship between sound pressure level and amplitude of signal, the relationship between the signal level ($s$) of a loudspeaker and sound pressure ($p$) is the second derivative: $$p=\frac{\partial^2 s}{\partial t^2}$$

And based on this Why does a microphone membrane only measure pressure and not particle velocity?, a microphone measures pressure.

So, if I put these two together, does this mean, if I:

  • play a wav file on a computer to a loudspeaker
  • record the output from the loudspeaker with a microphone
  • create a wav file from what the microphone records

then the recorded wav file will not have the original signal, but the second derivative of it (I supposed here that all the components are ideal, no losses, etc.)?

A second question: if I double the distance between the loudspeaker and microphone, what signal will the microphone record, what will be its amplitude related to the amplitude of the recording of the original position (I mean in the wav file)? $1/\sqrt 2$, $1/2$ or maybe $1/4$?

  • $\begingroup$ While this comes dangerously close to being an engineering question I think the core question is excellent. (Perhaps you could edit out the part about the wav file and concentrate on the relevant part: does a speaker precisely reproduce the signal recorded by a microphone). $\endgroup$ May 10, 2018 at 16:49
  • $\begingroup$ @SebastianRiese: yes, this is not strictly a physics question (if you can recommend a better to place to ask this, I'd happily move this question there). I added the wav thingy because I wanted to include the whole chain, because maybe, there is a "hidden" integrator somewhere in the chain, so in the end, a correct (original) signal will be recorded to the wav. $\endgroup$
    – geza
    May 10, 2018 at 17:05

1 Answer 1


If you follow the first post you reference to the end, you'll see this formula:

$p = \frac{\rho S_D}{2 \pi r} \frac{V_0 B\ell}{R_E M_\mathrm{ms}} \sin \omega t$

It suggests that, at least in theory, the pressure produced by a speaker is proportional to the input voltage.

A microphone could be viewed as a speaker in reverse. In fact, a speaker could be used as a microphone.

A microphone does respond to pressure: its sensitivity is typically expressed as the output voltage per unit of pressure, like dBV/Pa. So, ideally, the output voltage of a microphone is proportional to pressure.

So, there is no contradiction here: a speaker produces pressure proportional to the input voltage and the microphone converts that pressure to a proportional output voltage.

  • $\begingroup$ Thanks for the answer! Does this mean, that the idealized loudspeaker converts voltage to acceleration? And it is this way, because according to the Biot-Savart law, the strength of a magnetic field is proportional to the current in the wire, which is proportional to voltage in this case? $\endgroup$
    – geza
    May 10, 2018 at 18:32
  • $\begingroup$ @geza Yes, I believe so. $\endgroup$
    – V.F.
    May 10, 2018 at 18:43

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