If you have multiple waves of different frequencies, the interference from the different waves cause "beats".

(Animation from https://en.wikipedia.org/wiki/Group_velocity)


Let's say that a green dot in the above animation reaches your ear a few hundred times per second.

Is it possible to hear this phenomenon (wave groups occurring at frequencies in the audible range) as its own tone?

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    $\begingroup$ Perhaps the question you are asking is the same as Do we hear anything special when the beat frequency is in audible range, but the sounds producing the beats are not? That is a clearer question, although I think Pieter's answer is better. $\endgroup$ – sammy gerbil Oct 5 '18 at 18:49
  • $\begingroup$ @sammygerbil yes, that appears to be the same question, though worded differently enough that it didn’t appear in search $\endgroup$ – Daniel M. Oct 5 '18 at 19:25
  • $\begingroup$ I don't understand "if these things are audible, can you hear them?" By definition, when you call something audible, you can hear it. Perhaps you meant "if they create beats of an audible frequency, can you hear them?" $\endgroup$ – user191954 Oct 6 '18 at 5:38
  • $\begingroup$ @AaronStevens The OP has confirmed that the duplicate is the same question, and that does ask about ultrasound. I think it is necessary for this question to be clarified before more answers are posted. Ideally this should be done by the OP, but the OP has not responded quickly to my request. $\endgroup$ – sammy gerbil Oct 6 '18 at 9:45
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    $\begingroup$ @AaronStevens I modified the title to be, I think, clearer than it was before $\endgroup$ – Daniel M. Oct 6 '18 at 13:06

No, one cannot hear the actual beat frequency. For example, if both waves are ultrasonic and the difference in frequency is 440 Hz, you won't hear the A (unless some severe nonlinearities would come into play; edit: such nonlinear effects are at least 60 dB lower in sound pressure level).

When two ultrasonic waves are close in frequency, the amplitude goes up and down with the beat frequency. A microphone can show this on an oscilloscope. But the human ear does not hear the ultrasonic frequency. It is just silence varying in amplitude :)

(I know a physics textbook where this is wrong.)

Edit: in some cases the mind can perceive the pitch of a "missing fundamental". For example, when sine waves of 880 and 1320 Hz are played, the mind may perceive a tone of pitch A. This is a psychoacoustic phenomenon, exploited for example in the auditory illusion of an Escher's staircase.

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    $\begingroup$ @sammygerbil It is one way of explaining this to my students. Their textbook says that they should hear the beat frequency. Then I have them do the experiment. They see the signal on the oscilloscope, but all they hear is silence. "Silence varying in amplitude." So yes, a bit of a joke, but it helps them understand. $\endgroup$ – Pieter Oct 5 '18 at 18:59
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    $\begingroup$ @AaronStevens The point is that you cannot hear a beat of 1Hz if the carrier wave is outside of the audible range (ultrasonic). You can hear a beat when tuning the guitar because the carrier wave is always in the audible range. $\endgroup$ – sammy gerbil Oct 5 '18 at 22:05
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    $\begingroup$ actually yes. See my answer below. ATC depends on nonlinearity to demodulate the beat between two ultrasonic carriers sourced from a phased array. $\endgroup$ – docscience Oct 6 '18 at 0:25
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    $\begingroup$ One place that non-linear effects can occur is in speaker systems. So experiments to confirm this do rely on quality speakers with a flat frequency response at the ultrasonic frequencies being used. $\endgroup$ – Neil Slater Oct 6 '18 at 9:35
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    $\begingroup$ This answer makes the strong claim that nonlinearities are not present in the human auditory organs, without any evidence to support it. Human sensory organs are complicated systems and this type of property cannot be taken for granted without dedicated psychoacoustic experiments to support it. $\endgroup$ – Emilio Pisanty Oct 6 '18 at 11:52

Yes - American Technology Corporation, Woody Norris invented a phased array consisting of ultrasonic transducers; pairs that transmit two ultrasonic frequencies that are slightly different by a modulated sound frequency.

Demodulation of the audible signals from the ultrasonic carriers is accomplished either by nonlinear properties of air or by the two signals striking a surface such as a wall or the inside of your head! In any event the sound appears to occur virtually out of thin air.

These devices have been called hypersonic sound speakers or audio spotlights. Once in awhile you can find them for sale on EBay

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    $\begingroup$ To achieve such nonlinear effects one needs extreme intensities in the ultrasonic beams, achieved by focusing. Maybe of the order of one watt per square meter, which corresponds to "120 dB". Or even more? I do not think that this is what the OP asked about. $\endgroup$ – Pieter Oct 6 '18 at 7:29
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    $\begingroup$ Still, it's good to know that high intensities can lead to nonlinear effects which change normally no-go answer into a "yes-go" one. Similarly to optics, where materials normally can't absorb photons with energy 2× smaller than energy gap, but high intensities can lead to two-photon absorption. $\endgroup$ – Ruslan Oct 6 '18 at 8:59
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    $\begingroup$ I agree with Pieter: I don't think this non-linear phonomenon is what the OP was asking about. $\endgroup$ – sammy gerbil Oct 6 '18 at 9:35
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    $\begingroup$ @Pieter Can you provide the sources for the numbers you quote in this thread? As currently provided they look pulled out of thin air. $\endgroup$ – Emilio Pisanty Oct 6 '18 at 11:59
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    $\begingroup$ @Pieter actually it appears that there does exist nonlinearity in the inner ear at moderate sound pressures: see Non-linear Behavior of the Ear. $\endgroup$ – Ruslan Oct 6 '18 at 12:07

As always for anything involving biology, the answer is actually more complicated.

It is true that there is no "note" there at the beat frequency, in terms of Fourier series. But despite what is commonly stated in textbooks, the ear does not just do a Fourier transform.

In fact, the human ear does perceive differences in frequencies, and more generally certain linear combinations of frequencies, as actual tones. They are called combination tones, and a demo is here. As you can hear in the second clip, when two frequencies $f_1 < f_2$ are played, one hears tones at frequencies $f_2 - f_1$ (the difference tone) and at $2f_1 - f_2$ (the cubic difference tone), as well as some others. This is no small effect; these tones are several octaves below the original tones.

This would be impossible if the ear were a simple linear system, because there is no Fourier component at frequency $f_2 - f_1$ or $2f_1 - f_2$. But the ear is nonlinear, and its output is then subsequently processed by the brain, again in a nonlinear way. And it's well-known that the simplest thing nonlinearity can do is output linear combinations of the input tones; that is one of the cornerstones of nonlinear optics.

While the theory is not completely understood, almost everybody can hear the difference tones are there. However, in the case of extreme ultrasound, it's quite unlikely that you'd hear anything because an ultrasound wave can barely budge anything in your ear at all. If your ears are not sensitive enough to detect them in the first place, it's unlikely they would be able to output nonlinear combinations of them no matter how nonlinearly they process the sound.

  • $\begingroup$ I believe this is a phenomenon of perception in the mind. One may also hear binaural combination tones - combination tones that are audible when headphones present one sine frequency to one ear and the other frequency to the other ear. $\endgroup$ – Pieter Oct 6 '18 at 7:16
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    $\begingroup$ @Pieter The OP asked about hearing. Hearing is a phenomenon of perception. $\endgroup$ – knzhou Oct 6 '18 at 9:44

Hearing 'beats' at a frequency n, as in the above example, is not the same as hearing a note at that frequency. In the example you give, there is no actual note present at the lower frequency, i.e. the air is not being excited at that frequency. All you are hearing is an interference effect at frequency n. For example, if you were to convert that example waveform to the frequency domain (i.e. spectral analysis), you would see two higher-frequency spikes very close together, but there would be no spike present at the lower frequency n.

Your ear would hear and interpret the interference effect as the volume of the note increasing and decreasing at the frequency n. This effect can be used, for example, when tuning a guitar string - play two notes that are supposed to be the same on two different strings simultaneously and you will hear beats if they are slightly out of tune.

If you were to superimpose a lower-frequency note on top of a higher-frequency note (i.e. two notes played simultaneously), the waveform would look quite different (more like a high-frequency wave 'riding' a lower-frequency wave, as shown in the image below). In that case, your ear would hear the two different notes simultaneously.

enter image description here

  • $\begingroup$ The figure shows propagation in a very dispersive medium, where the group velocity is different from the phase velocity. This 'riding' does not happen in air. $\endgroup$ – Pieter Oct 5 '18 at 19:25
  • $\begingroup$ @Pieter I will add a image to illustrate what I mean by a higher-frequency wave 'riding' a lower-frequency one. My point is that the waveform of two frequencies superimposed would look very different to the 'beats' example given in the question. $\endgroup$ – Time4Tea Oct 5 '18 at 19:34
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    $\begingroup$ What is "the above example". The answers in StackExchange are not in any particular order. $\endgroup$ – JiK Oct 6 '18 at 18:36

The human ear detects sounds by having hairs with different fundamental frequencies; if an incoming frequency is sufficiently close to harmonic of the hair, the hair detects the sound. In essence, the ear performs an analog Fourier transform. While the graph of the beat looks like a sine wave, its dot product with a true sine wave is zero, thus it's not detectable.

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    $\begingroup$ I agree about the analog Fourier transform etc. But it is not due to the hair cells or the cilia themselves having different frequencies. It is because of their position on the basilar membrane in the conical tube of the cochlea. $\endgroup$ – Pieter Oct 5 '18 at 22:44
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    $\begingroup$ As Pieter said, the hairs have nothing to do with how we hear differentiate frequencies. $\endgroup$ – CramerTV Oct 5 '18 at 23:57

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