According to this answer (source), the time averaged power $P$ of an acoustic wave is:

$$P = \frac{1}{2} \mu v \omega^2 A^2$$

Where $\mu$ is mass density of the medium, $v$ is speed of sound in the medium, $\omega$ is the angular frequency of the wave, and $A$ is the amplitude of the wave.

Under the assumption that amplitude is in units of pressure (e.g. Pascals), this gives the following units for each variable (letters on right sides indicate units):

$$\begin{aligned} P &= \frac{kg \times m^2}{s^3} \text{ (Watts)} \\ \\ \mu &= \frac{kg}{m^3} \\ \\ v &= \frac{m}{s} \\ \\ \omega &= \frac{rad}{s} \\ \\ A &= \frac{kg}{m \times s^2} \text{ (Pascals)} \end{aligned}$$

But when attempting to balance the units, the units on the right side of that equation ends up as:

$$\begin{aligned} \frac{1}{2} \mu v \omega^2 A^2 &= \frac{kg}{m^3} \times \frac{m}{s} \times \frac{rad^2}{s^2} \times \frac{kg^2}{m^2 \times s^4} \\ \\ &= \frac{kg^3 \times rad^2}{m^4 \times s^7} \end{aligned}$$

These are clearly not units of power.

If I instead work backwards to figure out what amplitude needs to be:

$$\begin{aligned} \frac{kg \times m^2}{s^3} &= \frac{kg}{m^3} \times \frac{m}{s} \times \frac{rad^2}{s^2} \times \text{ ?}^2 \\ \\ \text{ ?}^2 &= \frac{m^4}{rad^2} \\ \\ \text{ ?} &= \frac{m^2}{rad} \end{aligned}$$

Then I get units of amplitude as square meters per radian, which is nonsense.

What's going on here?

Note: I'm concerned with acoustic pressure waves in e.g. air. The source, however, seems to be talking about oscillations on a string, which makes me wonder is this even the correct equation for pressure waves in a medium?

  • 2
    $\begingroup$ The source you linked even shows the amplitude is a displacement, so it has units of length. They show this immediately in an example under the equation for $P_{ave}$. $\endgroup$
    – Triatticus
    Feb 18 at 19:56
  • $\begingroup$ @Triatticus I see that now thanks. That still wouldn't balance the units in the equation, though. Also how can I convert sound pressure to a length...? I only know how to measure amplitude as a pressure (derived from a voltage from a microphone with a known frequency-dependent $V/Pa$). $\endgroup$
    – Jason C
    Feb 18 at 19:58
  • $\begingroup$ Is my question underspecified? I'm trying to use this for sound in air, rather than e.g. on a string. Also maybe my title is only a fraction of what I'm really trying to ask. But, I am too confused to put together a better title. I will try to think about it. $\endgroup$
    – Jason C
    Feb 18 at 20:01

1 Answer 1


I believe the power expression you list is a power density (just like the poynting vector in E&M). Thus $P$ has units of $W/m^2$, which enforces $A$ to have units of $m$ which is the standard unit of amplitude for a wave.

  • $\begingroup$ Thanks, and I think you're right, but now I'm wondering if this isn't the right equation for my situation. I'm working with pressure waves in a medium (e.g. air). The amplitude doesn't really have a length, or at least I don't have the knowledge to convert pressure to displacement. Usually electronic audio sensors (such as microphones) have specified frequency-dependent $V/Pa$ ratios, allowing you to calculate pressure from response voltage; so when you say "[length] is the standard unit of amplitude for a wave", it makes me think that I'm on the completely wrong track here. What am I missing? $\endgroup$
    – Jason C
    Feb 18 at 20:09
  • 2
    $\begingroup$ I believe the version of the power density equation you’re looking for is $p^2/(\rho v)$ where p is pressure in pascals. $\endgroup$
    – klippo
    Feb 18 at 20:20
  • 2
    $\begingroup$ Thanks. I think that's what I was looking for. You inspired me to search for en.wikipedia.org/wiki/Sound_power, it's all in there in some form or another. Sorry the original question was kind of unclear, I'm trying to muddle through this all on my own. Thanks for making sense of it! $\endgroup$
    – Jason C
    Feb 18 at 20:25

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