I was reading about how at how acoustic beats work.

If we combine two waves with frequencies $f_1$ and $f_2$ and unit amplitude, their combination is $$ \begin{align} A &= \cos\left(2\pi f_1x\right) + \cos\left(2\pi f_2x\right) \\[10px] &= 2\cos\left(2\pi \, \frac{f_1-f_2}{2}\,x\right)\cos\left(2\pi \, \frac{f_1+f_2}{2} \, x\right) \,. \end{align} $$

According to "Beat (acoustics)", Wikipedia:

Because the human ear is not sensitive to the phase of a sound, only its amplitude or intensity, only the magnitude of the envelope is heard.

So obviously the beat frequency is twice the envelope (since you're squaring it) and you get $$f_{\text{beat}} = f_1-f_2$$and not half that.

Now consider a regular cosine wave $A = \cos{\left(2\pi f_T\right)}$ with frequency $f_T$. Taking the magnitude (as Wikipedia says, i.e. by squaring $A$) gives you an audible frequency of $2f_T$... so do people hear frequencies as twice what they are in their amplitude wave?

EDIT: The answer is we DO perceive twice the frequency -- a sound wave that we define as having a frequency f will stimulate our ears with twice that frequency.

This frequency f is just a convenient name we give to the waves our machines make. This doesn't bother anyone as people can't hear in slow motion and i.e. count 200 'ticks' per second when playing a 100 Hz wave.

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    $\begingroup$ If they did, how could you measure it? $\endgroup$
    – user253751
    Jul 19, 2019 at 2:49
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    $\begingroup$ If humans always perceived them as double; they would never figure out that they're hearing double of what they should hear. This is the same principle like how our eyees work like a camera obscura (which inherently inverts the image) but humans automatically correct their perception so the world doesn't look upside down. $\endgroup$
    – Flater
    Jul 19, 2019 at 8:25
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    $\begingroup$ We don't "hear" frequencies. We perceive something close to a logarithmic fourier transform of what's moving in our ears, further processed and interpreted by our brains. $\endgroup$
    – OrangeDog
    Jul 19, 2019 at 10:32
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    $\begingroup$ @ThePhoton Probably the similarity of this question to the age old, "Is your blue the same as my blue?" $\endgroup$ Jul 19, 2019 at 16:39

5 Answers 5


Humans hear the correct perceptive signal for a sound wave of that frequency.

We really can't say much more than that. The psychology of acoustics are very complicated and could fill volumes.

It's closer to say we have cells which act resonant at a specific frequency. Our brain identifies which cells are resonating at any point in time, and constructs the signal from that. Our brains receive information that cell A or cell B is signalling. The association between those neural signals and frequencies is a learned response that we pick up early on, as an infant or perhaps even in the womb.

  • $\begingroup$ Yes. The frequencies are mapped to different distances in the cochlea. Only for low frequencies is there a relation between the action potentials and the phase of the wave. This plays a role in binaural direction sensing. $\endgroup$
    – user137289
    Jul 18, 2019 at 18:53
  • $\begingroup$ Ok now I understand that sound is really subjective to how our cells perceive it. I'm still a bit confused -- I know humans hear sound waves when there are compressions and expansions in our ears, and we can't tell the difference between the two. A sound wave of frequency 1 wave per second is defined as looking like a peak/trough sine wave (or a compression and then an expansion in a second). But since we can't tell the difference between compression and expansion, won't our ears feel this frequency "1" wave as happening twice per second (i.e. an actual frequency of "2" signals per second) $\endgroup$
    – Mondo Duke
    Jul 18, 2019 at 19:39
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    $\begingroup$ @MondoDuke A sine wave of 100 Hz causes movements of the basilar membrane at a different position than a sine wave of 200 Hz. Different hair cells are stimulated, different "threads" in the auditory nerve start firing. (But if you want to experience something weird, listen with headphones to binaural beats.) $\endgroup$
    – user137289
    Jul 18, 2019 at 19:55
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    $\begingroup$ We don't "sense" every cycle in the way you're thinking about it. A nerve fiber which is used in detecting 2kHz does not fire twice as fast as a nerve fiber which is used to detect 1kHz. Both fibers transmit something more akin to "here's how much power there is where my cells are at," and the cells are structured to do a fourier transform of sorts. $\endgroup$
    – Cort Ammon
    Jul 18, 2019 at 22:37
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    $\begingroup$ Aren't beats different than tones though? Isn't this what the OP is asking about? $\endgroup$ Jul 19, 2019 at 0:17

So obviously the audible frequency is twice the envelope

Sorry, that's wrong. If you play two tones (say 440 Hz and 267 Hz), you simply hear two tones at two different frequencies and you have two excitations at different spots on the basilar membrane and two different sets of nerves firing. You don't hear the envelope at all, they just sound like two steady-state tones.

"Beats" only happen when you have two frequencies that are VERY close together, say 237 Hz and 238 Hz. In this case, your ear can't resolve the frequency difference anymore but you hear a single tone at 237.5 Hz that's amplitude modulated at 1 Hz.

Taking the magnitude (as wikipedia says, i.e. by squaring A) gives you an audible frequency of 2fT

No. You can square the amplitude to estimate power or energy but there is no mechanism that would square the actual waveform. If you play 100 Hz, you hear 100 Hz, that's all there is to it.

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    $\begingroup$ Although, the apparent sine waves traced by the envelope have 1/2 Hz. Example $\endgroup$
    – Vaelus
    Jul 19, 2019 at 14:50
  • $\begingroup$ That's too strong of a statement to make. There's another theory of hearing, the temporal theory, where the neurons really do record the frequency directly. I'm under the impression that neither the temporal theory nor yours (called the "place" theory) can explain all observations; the real, messy process of hearing might use both. $\endgroup$
    – knzhou
    Jul 20, 2019 at 14:22

The human perception of a wave at frequency $f$ is the human perception of a wave at frequency $f$. There is no "objective" qualia for frequency $f$ other than what people perceive, so it's nonsensical to ask whether people, when they hear $f$, perceive $2f$; there is no meaning to "perceive $2f$" other than "experience the qualia associated with $2f$", and clearly when someone hears $f$, they experience that qualia associated with $f$, not $2f$.

The human ear basically is a device for detecting components of the Fourier transform of sound. The reason that $f_2-f_1$ dominates with beats is that if $f_2+f_1$ is high enough, then the $f_2-f_1$ component will not be significantly affected by multiplying by a $f_2+f_1$ wave.

  • $\begingroup$ This is true for the perception of pitch for a signal constant-amplitude signal. But it isn't true for beats, which operate through a very different mechanism. With beats, you're heading the modulation in the envelope amplitude, not the phase oscillation. And this modulation is over slow enough time scales that you really can directly measure it, e.g. with a stopwatch. You hear beats as a periodic oscillation in volume, not as a pitch. $\endgroup$
    – tparker
    Dec 12, 2021 at 1:41

Your intuition is right. It seems you might have missed this statement in the same Wikipedia article which confirms what you're asking:

Therefore, subjectively, the frequency of the envelope seems to have twice the frequency of the modulating cosine, which means the audible beat frequency is: $$f_{\text{beat}} = f_1-f_2$$

Basically the wavelength of a beat, as far as hearing is concerned, is the duration between successive amplitude maxima, and not the abstract modulating cosine wave which has twice that length.

  • $\begingroup$ I think we both said the same things lol. in other words, we're on the same wavelength $\endgroup$
    – Mondo Duke
    Aug 26, 2020 at 20:55

The human ear is only sensitive to the amplitude in the sense that you can't tell apart $sin(t)$ and $sin(t+\phi)$. It doesn't mean you cannot tell apart $sin(t)$ and $sin^2(t)$: the latter will be heard as twice the frequency at half the volume.

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    $\begingroup$ @Jasper I think he meant sensitive to amplitude and frequency but not to phase (as the rest of the sentence suggests). $\endgroup$ Jul 21, 2019 at 0:11

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