Microphones (e.g. condenser microphone) are assumed to have a voltage output proportional to the sound pressure at the diaphragm.

If the operating principle is that the voltage output is proportional to the displacement of the thin foil that is the diaphragm, then why does the particle velocity of the sound wave not affect the displacement of the diaphragm and hence the measurement?

I know that in acoustics the particle velocity and the acoustic pressure do not have a fixed relationship (except for the case of plane waves), and hence in the near-field of a source, a sound intensity measurement must be taken (uses two microphones to estimate particle velocity).

It is obvious that the force from the acoustic pressure at the film will cause a velocity (hence displacement) of the film, but what about the contribution from particle velocity of the fluid transferring momentum to the film?

  • $\begingroup$ Are high pressure and high particle velocity mutually exclusive? $\endgroup$
    – BMS
    Commented Mar 25, 2014 at 18:28
  • $\begingroup$ @BMS: yes (in the fluid). e.g. destructive interference of two waves will reduce the pressure whilst increasing the particle velocity, or in the near-field of a simple point source the pressure and velocity do not have a simple relationship. $\endgroup$
    – xyz
    Commented Mar 25, 2014 at 18:47

2 Answers 2


Think about the definition of pressure: $$P=\frac{F}{A}$$ Now, let's consider the definition of a force. $$F=\frac{dp}{dt}=m\frac{dv}{dt}$$ Hence, for a given area and particle mass, the pressure is a function of the velocity: $$P=\frac{m}{A}\frac{dv}{dt} $$

  • $\begingroup$ Sure, the acoustic pressure causes this velocity at the membrane, but why is there no contribution to this velocity from the velocity of the air particles impacting the membrane too? I have updated my question to try and explain my confusion. $\endgroup$
    – xyz
    Commented Mar 25, 2014 at 18:52
  • 2
    $\begingroup$ @James you are mixing two frameworks here. When one is talking of "particles" one is in a different framework where pressure is defined. One can talk of the velocity of the particles in the air in the framework of kinetic theory of gases, and there one finds that pressure is the motion of particles hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kinthe.html $\endgroup$
    – anna v
    Commented Mar 25, 2014 at 19:21
  • $\begingroup$ @James the velocity I'm talking about is that of the particles. I'm working in the framework of the kinetic theory of gases, as anna v pointed out correctly. The particles hit the membrane and it therefore experiences a pressure. $\endgroup$
    – Danu
    Commented Mar 25, 2014 at 20:00
  • $\begingroup$ But at a boundary there is only pressure and no velocity. The velocity of the microphone diaphragm for a given pressure depends on its acoustic impedance, which can be arbitrarily large like the input impedance of a circuit. $\endgroup$
    – endolith
    Commented May 13, 2016 at 0:39

If one uses a fan to blow air on a fixed disk, the air pressure on the side of the disk facing the fan will be higher than on the reverse. The relationship between the movement and pressure will depend upon a variety of factors including the size of the disk, angle at which the air is blowing, etc. The net force applied to the disk, however, will be proportional to the difference in pressure on the two sides, regardless of what pattern of air movement created it.

If one had a microphone whose frequency response were completely flat with regard to pressure, then a 2,500Hz signal would represent about 1% as much air movement as a 250Hz signal at the same pressure level, which would in turn be about 1% as much air movement as a 25Hz signal. Thus, a microphone which detected air movement rather than pressure would have, by conventional standards, an exceptionally-bass-heavy frequency response, with a 12dB/octave drop-off. If a recording level was set so that a 65.5Hz tone at a certain pressure level would be full-scale, a 2096Hz tone at the same pressure level would be 0.1% of full scale. Circuitry would thus have to have a very good signal-to-noise ratio by modern standards in order to achieve even moderate fidelity.


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