Mixing
If two ultrasound frequencies create a beat in the audible frequency range, can you hear it?
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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 the waves which interfere are outside of the audible frequency range, and the beats are within at the audible frequency range. For example, a green dot in the above animation reaches your ear a few hundred times per second.
Is it possible to hear the beats?
waves acoustics interference biophysics
edited 18 mins ago
sammy gerbil
21.6k42355
asked 15 hours ago
Daniel M.
1786
 |Â
show 6 more comments
up vote
13
down vote
favorite
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 the waves which interfere are outside of the audible frequency range, and the beats are within at the audible frequency range. For example, a green dot in the above animation reaches your ear a few hundred times per second.
Is it possible to hear the beats?
waves acoustics interference biophysics
edited 18 mins ago
sammy gerbil
21.6k42355
asked 15 hours ago
Daniel M.
1786
Are you asking if you can hear the beats? If so, then yes you can.
– Aaron Stevens
15 hours ago
1
@sammygerbil Hearing beats = the sound fades in and out at n times a second. Hearing wave groups themselves = hearing the tone for n Hz (where n is the number of wave groups per second)
– Daniel M.
15 hours ago
4
'Hearing' something simply means that the air pressure vibrates your eardrum. In the above example there is no pressure or 'sound' at 440Hz. The absence of a sound at 440Hz is not going to let you hear an A.
– CramerTV
15 hours ago
4
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.
– sammy gerbil
14 hours ago
2
@DanielM Please would you edit your question to make the meaning clearer. Then we can vote to close the other question as a duplicate of yours, because the better answers are with this question.
– sammy gerbil
11 hours ago
 |Â
show 6 more comments
up vote
13
down vote
favorite
up vote
13
down vote
favorite
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 the waves which interfere are outside of the audible frequency range, and the beats are within at the audible frequency range. For example, a green dot in the above animation reaches your ear a few hundred times per second.
Is it possible to hear the beats?
waves acoustics interference biophysics
edited 18 mins ago
sammy gerbil
21.6k42355
asked 15 hours ago
Daniel M.
1786
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 the waves which interfere are outside of the audible frequency range, and the beats are within at the audible frequency range. For example, a green dot in the above animation reaches your ear a few hundred times per second.
Is it possible to hear the beats?
waves acoustics interference biophysics
waves acoustics interference biophysics
edited 18 mins ago
sammy gerbil
21.6k42355
asked 15 hours ago
Daniel M.
1786
edited 18 mins ago
sammy gerbil
21.6k42355
asked 15 hours ago
Daniel M.
1786
edited 18 mins ago
sammy gerbil
21.6k42355
edited 18 mins ago
sammy gerbil
21.6k42355
edited 18 mins ago
sammy gerbil
21.6k42355
21.6k42355
asked 15 hours ago
Daniel M.
1786
asked 15 hours ago
Daniel M.
1786
asked 15 hours ago
Daniel M.
1786
1786
Are you asking if you can hear the beats? If so, then yes you can.
– Aaron Stevens
15 hours ago
1
@sammygerbil Hearing beats = the sound fades in and out at n times a second. Hearing wave groups themselves = hearing the tone for n Hz (where n is the number of wave groups per second)
– Daniel M.
15 hours ago
4
'Hearing' something simply means that the air pressure vibrates your eardrum. In the above example there is no pressure or 'sound' at 440Hz. The absence of a sound at 440Hz is not going to let you hear an A.
– CramerTV
15 hours ago
4
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.
– sammy gerbil
14 hours ago
2
@DanielM Please would you edit your question to make the meaning clearer. Then we can vote to close the other question as a duplicate of yours, because the better answers are with this question.
– sammy gerbil
11 hours ago
 |Â
show 6 more comments
Are you asking if you can hear the beats? If so, then yes you can.
– Aaron Stevens
15 hours ago
1
@sammygerbil Hearing beats = the sound fades in and out at n times a second. Hearing wave groups themselves = hearing the tone for n Hz (where n is the number of wave groups per second)
– Daniel M.
15 hours ago
4
'Hearing' something simply means that the air pressure vibrates your eardrum. In the above example there is no pressure or 'sound' at 440Hz. The absence of a sound at 440Hz is not going to let you hear an A.
– CramerTV
15 hours ago
4
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.
– sammy gerbil
14 hours ago
2
@DanielM Please would you edit your question to make the meaning clearer. Then we can vote to close the other question as a duplicate of yours, because the better answers are with this question.
– sammy gerbil
11 hours ago
Are you asking if you can hear the beats? If so, then yes you can.
– Aaron Stevens
15 hours ago
Are you asking if you can hear the beats? If so, then yes you can.
– Aaron Stevens
15 hours ago
1
1
@sammygerbil Hearing beats = the sound fades in and out at n times a second. Hearing wave groups themselves = hearing the tone for n Hz (where n is the number of wave groups per second)
– Daniel M.
15 hours ago
@sammygerbil Hearing beats = the sound fades in and out at n times a second. Hearing wave groups themselves = hearing the tone for n Hz (where n is the number of wave groups per second)
– Daniel M.
15 hours ago
4
4
'Hearing' something simply means that the air pressure vibrates your eardrum. In the above example there is no pressure or 'sound' at 440Hz. The absence of a sound at 440Hz is not going to let you hear an A.
– CramerTV
15 hours ago
'Hearing' something simply means that the air pressure vibrates your eardrum. In the above example there is no pressure or 'sound' at 440Hz. The absence of a sound at 440Hz is not going to let you hear an A.
– CramerTV
15 hours ago
4
4
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.
– sammy gerbil
14 hours ago
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.
– sammy gerbil
14 hours ago
2
2
@DanielM Please would you edit your question to make the meaning clearer. Then we can vote to close the other question as a duplicate of yours, because the better answers are with this question.
– sammy gerbil
11 hours ago
@DanielM Please would you edit your question to make the meaning clearer. Then we can vote to close the other question as a duplicate of yours, because the better answers are with this question.
– sammy gerbil
11 hours ago
 |Â
show 6 more comments
5 Answers
5
active
oldest
votes
up vote
7
down vote
accepted
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).
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.
answered 15 hours ago
Pieter
5,94231026
1
What is "silence varying in amplitude"? Or am I missing your joke?
– sammy gerbil
14 hours ago
1
@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.
– Pieter
14 hours ago
What do you mean by not being able to hear beat frequencies? Do you mean you won't hear an audible pitch? Because you can definitely hear the beating depending on the frequency difference. This is how people tune instruments based on a certain frequency.
– Aaron Stevens
14 hours ago
4
@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.
– sammy gerbil
11 hours ago
1
actually yes. See my answer below. ATC depends on nonlinearity to demodulate the beat between two ultrasonic carriers sourced from a phased array.
– docscience
9 hours ago
 |Â
show 5 more comments
up vote
6
down vote
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
answered 9 hours ago
docscience
8,35411748
3
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.
– Pieter
2 hours ago
1
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.
– Ruslan
42 mins ago
I agree with Pieter: I don't think this non-linear phonomenon is what the OP was asking about.
– sammy gerbil
7 mins ago
add a comment |Â
up vote
4
down vote
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.
answered 10 hours ago
knzhou
36.4k9102174
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.
– Pieter
2 hours ago
add a comment |Â
up vote
3
down vote
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.
answered 14 hours ago
Time4Tea
2,201930
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.
– Pieter
14 hours ago
@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.
– Time4Tea
14 hours ago
add a comment |Â
up vote
0
down vote
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.
answered 11 hours ago
Acccumulation
1,53319
5
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.
– Pieter
10 hours ago
2
As Pieter said, the hairs have nothing to do with how we hear differentiate frequencies.
– CramerTV
9 hours ago
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
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).
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.
answered 15 hours ago
Pieter
5,94231026
1
What is "silence varying in amplitude"? Or am I missing your joke?
– sammy gerbil
14 hours ago
1
@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.
– Pieter
14 hours ago
What do you mean by not being able to hear beat frequencies? Do you mean you won't hear an audible pitch? Because you can definitely hear the beating depending on the frequency difference. This is how people tune instruments based on a certain frequency.
– Aaron Stevens
14 hours ago
4
@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.
– sammy gerbil
11 hours ago
1
actually yes. See my answer below. ATC depends on nonlinearity to demodulate the beat between two ultrasonic carriers sourced from a phased array.
– docscience
9 hours ago
 |Â
show 5 more comments
up vote
7
down vote
accepted
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).
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.
answered 15 hours ago
Pieter
5,94231026
1
What is "silence varying in amplitude"? Or am I missing your joke?
– sammy gerbil
14 hours ago
1
@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.
– Pieter
14 hours ago
What do you mean by not being able to hear beat frequencies? Do you mean you won't hear an audible pitch? Because you can definitely hear the beating depending on the frequency difference. This is how people tune instruments based on a certain frequency.
– Aaron Stevens
14 hours ago
4
@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.
– sammy gerbil
11 hours ago
1
actually yes. See my answer below. ATC depends on nonlinearity to demodulate the beat between two ultrasonic carriers sourced from a phased array.
– docscience
9 hours ago
 |Â
show 5 more comments
up vote
7
down vote
accepted
up vote
7
down vote
accepted
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).
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.
answered 15 hours ago
Pieter
5,94231026
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).
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.
answered 15 hours ago
Pieter
5,94231026
edited 10 hours ago
answered 15 hours ago
Pieter
5,94231026
answered 15 hours ago
Pieter
5,94231026
answered 15 hours ago
Pieter
5,94231026
5,94231026
1
What is "silence varying in amplitude"? Or am I missing your joke?
– sammy gerbil
14 hours ago
1
@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.
– Pieter
14 hours ago
What do you mean by not being able to hear beat frequencies? Do you mean you won't hear an audible pitch? Because you can definitely hear the beating depending on the frequency difference. This is how people tune instruments based on a certain frequency.
– Aaron Stevens
14 hours ago
4
@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.
– sammy gerbil
11 hours ago
1
actually yes. See my answer below. ATC depends on nonlinearity to demodulate the beat between two ultrasonic carriers sourced from a phased array.
– docscience
9 hours ago
 |Â
show 5 more comments
1
What is "silence varying in amplitude"? Or am I missing your joke?
– sammy gerbil
14 hours ago
1
@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.
– Pieter
14 hours ago
What do you mean by not being able to hear beat frequencies? Do you mean you won't hear an audible pitch? Because you can definitely hear the beating depending on the frequency difference. This is how people tune instruments based on a certain frequency.
– Aaron Stevens
14 hours ago
4
@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.
– sammy gerbil
11 hours ago
1
actually yes. See my answer below. ATC depends on nonlinearity to demodulate the beat between two ultrasonic carriers sourced from a phased array.
– docscience
9 hours ago
1
1
What is "silence varying in amplitude"? Or am I missing your joke?
– sammy gerbil
14 hours ago
What is "silence varying in amplitude"? Or am I missing your joke?
– sammy gerbil
14 hours ago
1
1
@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.
– Pieter
14 hours ago
@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.
– Pieter
14 hours ago
What do you mean by not being able to hear beat frequencies? Do you mean you won't hear an audible pitch? Because you can definitely hear the beating depending on the frequency difference. This is how people tune instruments based on a certain frequency.
– Aaron Stevens
14 hours ago
What do you mean by not being able to hear beat frequencies? Do you mean you won't hear an audible pitch? Because you can definitely hear the beating depending on the frequency difference. This is how people tune instruments based on a certain frequency.
– Aaron Stevens
14 hours ago
4
4
@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.
– sammy gerbil
11 hours ago
@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.
– sammy gerbil
11 hours ago
1
1
actually yes. See my answer below. ATC depends on nonlinearity to demodulate the beat between two ultrasonic carriers sourced from a phased array.
– docscience
9 hours ago
actually yes. See my answer below. ATC depends on nonlinearity to demodulate the beat between two ultrasonic carriers sourced from a phased array.
– docscience
9 hours ago
 |Â
show 5 more comments
up vote
6
down vote
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
answered 9 hours ago
docscience
8,35411748
3
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.
– Pieter
2 hours ago
1
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.
– Ruslan
42 mins ago
I agree with Pieter: I don't think this non-linear phonomenon is what the OP was asking about.
– sammy gerbil
7 mins ago
add a comment |Â
up vote
6
down vote
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
answered 9 hours ago
docscience
8,35411748
3
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.
– Pieter
2 hours ago
1
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.
– Ruslan
42 mins ago
I agree with Pieter: I don't think this non-linear phonomenon is what the OP was asking about.
– sammy gerbil
7 mins ago
add a comment |Â
up vote
6
down vote
up vote
6
down vote
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
answered 9 hours ago
docscience
8,35411748
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
answered 9 hours ago
docscience
8,35411748
edited 9 hours ago
answered 9 hours ago
docscience
8,35411748
answered 9 hours ago
docscience
8,35411748
answered 9 hours ago
docscience
8,35411748
8,35411748
3
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.
– Pieter
2 hours ago
1
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.
– Ruslan
42 mins ago
I agree with Pieter: I don't think this non-linear phonomenon is what the OP was asking about.
– sammy gerbil
7 mins ago
add a comment |Â
3
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.
– Pieter
2 hours ago
1
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.
– Ruslan
42 mins ago
I agree with Pieter: I don't think this non-linear phonomenon is what the OP was asking about.
– sammy gerbil
7 mins ago
3
3
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.
– Pieter
2 hours ago
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.
– Pieter
2 hours ago
1
1
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.
– Ruslan
42 mins ago
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.
– Ruslan
42 mins ago
I agree with Pieter: I don't think this non-linear phonomenon is what the OP was asking about.
– sammy gerbil
7 mins ago
I agree with Pieter: I don't think this non-linear phonomenon is what the OP was asking about.
– sammy gerbil
7 mins ago
add a comment |Â
up vote
4
down vote
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.
answered 10 hours ago
knzhou
36.4k9102174
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.
– Pieter
2 hours ago
add a comment |Â
up vote
4
down vote
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.
answered 10 hours ago
knzhou
36.4k9102174
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.
– Pieter
2 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
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.
answered 10 hours ago
knzhou
36.4k9102174
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.
answered 10 hours ago
knzhou
36.4k9102174
answered 10 hours ago
knzhou
36.4k9102174
answered 10 hours ago
knzhou
36.4k9102174
answered 10 hours ago
knzhou
36.4k9102174
36.4k9102174
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.
– Pieter
2 hours ago
add a comment |Â
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.
– Pieter
2 hours ago
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.
– Pieter
2 hours ago
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.
– Pieter
2 hours ago
add a comment |Â
up vote
3
down vote
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.
answered 14 hours ago
Time4Tea
2,201930
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.
– Pieter
14 hours ago
@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.
– Time4Tea
14 hours ago
add a comment |Â
up vote
3
down vote
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.
answered 14 hours ago
Time4Tea
2,201930
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.
– Pieter
14 hours ago
@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.
– Time4Tea
14 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
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.
answered 14 hours ago
Time4Tea
2,201930
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.
answered 14 hours ago
Time4Tea
2,201930
edited 14 hours ago
answered 14 hours ago
Time4Tea
2,201930
answered 14 hours ago
Time4Tea
2,201930
answered 14 hours ago
Time4Tea
2,201930
2,201930
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.
– Pieter
14 hours ago
@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.
– Time4Tea
14 hours ago
add a comment |Â
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.
– Pieter
14 hours ago
@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.
– Time4Tea
14 hours ago
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.
– Pieter
14 hours ago
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.
– Pieter
14 hours ago
@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.
– Time4Tea
14 hours ago
@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.
– Time4Tea
14 hours ago
add a comment |Â
up vote
0
down vote
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.
answered 11 hours ago
Acccumulation
1,53319
5
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.
– Pieter
10 hours ago
2
As Pieter said, the hairs have nothing to do with how we hear differentiate frequencies.
– CramerTV
9 hours ago
add a comment |Â
up vote
0
down vote
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.
answered 11 hours ago
Acccumulation
1,53319
5
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.
– Pieter
10 hours ago
2
As Pieter said, the hairs have nothing to do with how we hear differentiate frequencies.
– CramerTV
9 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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.
answered 11 hours ago
Acccumulation
1,53319
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.
answered 11 hours ago
Acccumulation
1,53319
answered 11 hours ago
Acccumulation
1,53319
answered 11 hours ago
Acccumulation
1,53319
answered 11 hours ago
Acccumulation
1,53319
1,53319
5
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.
– Pieter
10 hours ago
2
As Pieter said, the hairs have nothing to do with how we hear differentiate frequencies.
– CramerTV
9 hours ago
add a comment |Â
5
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.
– Pieter
10 hours ago
2
As Pieter said, the hairs have nothing to do with how we hear differentiate frequencies.
– CramerTV
9 hours ago
5
5
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.
– Pieter
10 hours ago
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.
– Pieter
10 hours ago
2
2
As Pieter said, the hairs have nothing to do with how we hear differentiate frequencies.
– CramerTV
9 hours ago
As Pieter said, the hairs have nothing to do with how we hear differentiate frequencies.
– CramerTV
9 hours ago
add a comment |Â
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Are you asking if you can hear the beats? If so, then yes you can.
– Aaron Stevens
15 hours ago
1
@sammygerbil Hearing beats = the sound fades in and out at n times a second. Hearing wave groups themselves = hearing the tone for n Hz (where n is the number of wave groups per second)
– Daniel M.
15 hours ago
4
'Hearing' something simply means that the air pressure vibrates your eardrum. In the above example there is no pressure or 'sound' at 440Hz. The absence of a sound at 440Hz is not going to let you hear an A.
– CramerTV
15 hours ago
4
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.
– sammy gerbil
14 hours ago
2
@DanielM Please would you edit your question to make the meaning clearer. Then we can vote to close the other question as a duplicate of yours, because the better answers are with this question.
– sammy gerbil
11 hours ago