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Sound waves in different octaves

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Sound waves in different octaves


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2












$begingroup$


How do the sound waves compare between different octaves? Why is it that you can play a low c and a c, and they are at different frequencies, but the same tone? An octave above a note is the fourth overtone, I'm assuming, so would the slopes of the waves align? You can still tune to these notes between octaves, and the beats can still be heard if out of tune. My AP physics teacher was unable to answer why.










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Ava Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    The word tone refers to a sound or note, or even an out of tune note. The higher c is a different tone than the middle c. Octaves are double (or half) the frequency of any note. Middle c is 440Hz, its octave is 880Hz ( another c). Also 220Hz, see wikipedia.
    $endgroup$
    – PhysicsDave
    4 hours ago










  • $begingroup$
    I feel like this a question more about human perception rather than physics. The physics just says that (what we call) octaves are at different frequencies. The similarity, pleasantness, etc. of sounds when compared is purely based on our perception.
    $endgroup$
    – Aaron Stevens
    4 hours ago






  • 1




    $begingroup$
    @PhysicsDave Middle c is at 262 hertz (in current standard tuning where a' is att 440 hertz).
    $endgroup$
    – Pieter
    3 hours ago


















2












$begingroup$


How do the sound waves compare between different octaves? Why is it that you can play a low c and a c, and they are at different frequencies, but the same tone? An octave above a note is the fourth overtone, I'm assuming, so would the slopes of the waves align? You can still tune to these notes between octaves, and the beats can still be heard if out of tune. My AP physics teacher was unable to answer why.










share|cite|improve this question









New contributor




Ava Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    The word tone refers to a sound or note, or even an out of tune note. The higher c is a different tone than the middle c. Octaves are double (or half) the frequency of any note. Middle c is 440Hz, its octave is 880Hz ( another c). Also 220Hz, see wikipedia.
    $endgroup$
    – PhysicsDave
    4 hours ago










  • $begingroup$
    I feel like this a question more about human perception rather than physics. The physics just says that (what we call) octaves are at different frequencies. The similarity, pleasantness, etc. of sounds when compared is purely based on our perception.
    $endgroup$
    – Aaron Stevens
    4 hours ago






  • 1




    $begingroup$
    @PhysicsDave Middle c is at 262 hertz (in current standard tuning where a' is att 440 hertz).
    $endgroup$
    – Pieter
    3 hours ago
















2












2








2





$begingroup$


How do the sound waves compare between different octaves? Why is it that you can play a low c and a c, and they are at different frequencies, but the same tone? An octave above a note is the fourth overtone, I'm assuming, so would the slopes of the waves align? You can still tune to these notes between octaves, and the beats can still be heard if out of tune. My AP physics teacher was unable to answer why.










share|cite|improve this question









New contributor




Ava Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




How do the sound waves compare between different octaves? Why is it that you can play a low c and a c, and they are at different frequencies, but the same tone? An octave above a note is the fourth overtone, I'm assuming, so would the slopes of the waves align? You can still tune to these notes between octaves, and the beats can still be heard if out of tune. My AP physics teacher was unable to answer why.







waves acoustics frequency






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New contributor




Ava Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Ava Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Ben Crowell

53k6164310




53k6164310






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asked 5 hours ago









Ava Miller Ava Miller

112




112




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New contributor





Ava Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Ava Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    The word tone refers to a sound or note, or even an out of tune note. The higher c is a different tone than the middle c. Octaves are double (or half) the frequency of any note. Middle c is 440Hz, its octave is 880Hz ( another c). Also 220Hz, see wikipedia.
    $endgroup$
    – PhysicsDave
    4 hours ago










  • $begingroup$
    I feel like this a question more about human perception rather than physics. The physics just says that (what we call) octaves are at different frequencies. The similarity, pleasantness, etc. of sounds when compared is purely based on our perception.
    $endgroup$
    – Aaron Stevens
    4 hours ago






  • 1




    $begingroup$
    @PhysicsDave Middle c is at 262 hertz (in current standard tuning where a' is att 440 hertz).
    $endgroup$
    – Pieter
    3 hours ago
















  • 1




    $begingroup$
    The word tone refers to a sound or note, or even an out of tune note. The higher c is a different tone than the middle c. Octaves are double (or half) the frequency of any note. Middle c is 440Hz, its octave is 880Hz ( another c). Also 220Hz, see wikipedia.
    $endgroup$
    – PhysicsDave
    4 hours ago










  • $begingroup$
    I feel like this a question more about human perception rather than physics. The physics just says that (what we call) octaves are at different frequencies. The similarity, pleasantness, etc. of sounds when compared is purely based on our perception.
    $endgroup$
    – Aaron Stevens
    4 hours ago






  • 1




    $begingroup$
    @PhysicsDave Middle c is at 262 hertz (in current standard tuning where a' is att 440 hertz).
    $endgroup$
    – Pieter
    3 hours ago










1




1




$begingroup$
The word tone refers to a sound or note, or even an out of tune note. The higher c is a different tone than the middle c. Octaves are double (or half) the frequency of any note. Middle c is 440Hz, its octave is 880Hz ( another c). Also 220Hz, see wikipedia.
$endgroup$
– PhysicsDave
4 hours ago




$begingroup$
The word tone refers to a sound or note, or even an out of tune note. The higher c is a different tone than the middle c. Octaves are double (or half) the frequency of any note. Middle c is 440Hz, its octave is 880Hz ( another c). Also 220Hz, see wikipedia.
$endgroup$
– PhysicsDave
4 hours ago












$begingroup$
I feel like this a question more about human perception rather than physics. The physics just says that (what we call) octaves are at different frequencies. The similarity, pleasantness, etc. of sounds when compared is purely based on our perception.
$endgroup$
– Aaron Stevens
4 hours ago




$begingroup$
I feel like this a question more about human perception rather than physics. The physics just says that (what we call) octaves are at different frequencies. The similarity, pleasantness, etc. of sounds when compared is purely based on our perception.
$endgroup$
– Aaron Stevens
4 hours ago




1




1




$begingroup$
@PhysicsDave Middle c is at 262 hertz (in current standard tuning where a' is att 440 hertz).
$endgroup$
– Pieter
3 hours ago






$begingroup$
@PhysicsDave Middle c is at 262 hertz (in current standard tuning where a' is att 440 hertz).
$endgroup$
– Pieter
3 hours ago












4 Answers
4






active

oldest

votes


















3












$begingroup$

The phenomenon you're asking about is known as octave equivalence. It's a hard-wired thing in the human ear-brain system. (We know it's hard-wired because it's present without musical training and is true cross-culturally.) Notes that differ in frequency by a factor of 2 (or a power of 2) are perceptually similar, and may be mistaken for one another, even by trained musicians. So this is really a fact about the ear-brain system (i.e., psychology and physiology) rather than physics, although it does relate to a physical property.



Looking through the index entries on this topic in Diana Deutsch, The psychology of music, 3rd ed., 2013, I don't see anything about any physical, neurological, or evolutionary explanation of octave equivalence. That may mean that nobody has such an explanation, or just that it wasn't something that Deutsch wanted to get into in this anthology of articles. It seems likely that it has at least some physical basis, because periodic tones usually have their first two frequencies (fundamental and first harmonic) in the ratio of two to one.




An octave above a note is the fourth overtone, I'm assuming




No, it's the first overtone.




How do the sound waves compare between different octaves?




Pitch perception can be complicated in some cases, but essentially our sense of pitch is normally based on the period of the wave. So notes differing by an octave in pitch have waveforms that are related by the fact that their periods are in a 2 to 1 ratio.




You can still tune to these notes between octaves, and the beats can still be heard if out of tune.




The beats would be beats between the first harmonic of the lower note and the fundamental of the higher note.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If you think about it, the human ear contains thousands (if not millions) of tiny hairs which vibrate at different frequencies. The nervous system detects the vibration as "sound". But, as with guitar strings, a hair that is "tuned" to a given frequency will tend to vibrate at frequencies twice as high, or, to a lesser extent, half as high. To avoid befuddling the poor human, the nervous system somehow combines the various related harmonics into a single sensation, with high or low being a separate sensation.
    $endgroup$
    – Hot Licks
    15 mins ago










  • $begingroup$
    @HotLicks: Normally a resonator does not respond strongly to a driving force at twice its resonant frequency, so it's far from obvious that the mechanism you propose should occur.
    $endgroup$
    – Ben Crowell
    8 mins ago










  • $begingroup$
    You're assuming linearity.
    $endgroup$
    – Hot Licks
    6 mins ago



















2












$begingroup$

I basically agree with the answers of Ben Crowell and of Pieter. The fact that we hear the octave as a pleasant pure interval is hard-wired in the human hearing system. As Pieter explained, musical instruments produce tones with many overtones of different intensities, where usually the frequency assigned to the tone is the lowest, and the overtones are integer multiples of it.



The point I can add to these previous answers gives a hint, why the octave is a very special interval sounding most pure. The point is that if you produce a perfect octave on a musical instrument, the frequencies of all the overtones of the higher note will exactly line up with frequencies of the overtones of the lower note. This means that any beating in the resulting sound is reduced to a minimum.



The situation is already different for a perfect fifth. In this case only every second overtone of the higher note lines up with an overtone of the lower note.



The situation gets even worse for the other intervals. For example, for the minor third, only every fifth overtone lines up, and other overtones start to generate beatings. For dissonances, like the minor second, none of the overtones line up, and severe beatings arise from the overtones, resulting in a 'rough' perceived sound.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    The human voice and most musical instruments produce tones with many overtones. The sound pressure is periodic with a period $T$, but does not vary like a sine wave at all. Instead, it can be described by a Fourier series. That is the sum of sines that are integer multiples of the fundamental frequency $f=1/T$.



    So when a singer sings with a fundamental at 220 hertz (an a), that note includes overtones at 440 hertz (a'), 660 hertz (e"), 880 hertz (a") etcetera. This is why beats can be heard between two instruments that play an a and an a' slightly out of tune.




    would the slopes of the waves align?




    The ear does a kind of Fourier analysis of the input waveform, where different frequencies are mapped to different parts of the cochlea and to different nerves going to the brain. Our hearing is almost insensitive to the relative phase between the frequency components.



    One can look at the frequency content of sounds with spectrogram apps on mobile phones or on https://musiclab.chromeexperiments.com/






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      This doesn't answer the question, which is about octave identification.
      $endgroup$
      – Ben Crowell
      3 hours ago










    • $begingroup$
      @BenCrowell The OP also wrote something about beats.
      $endgroup$
      – Pieter
      3 hours ago



















    0












    $begingroup$

    The note $A4$ is $440Hz$. The note $A2$ is $220Hz$. They are one octave apart. We know that $s=lambda f$ where $s$ is the speed of sound in air, $lambda$ is wavelength, and $f$ is wave frequency. In terms of wavelength we have $$lambda=frac{s}{f}$$ So when we double the frequency going from $A2$ to $A4$, the wavelength is cut in half. Changing the frequency (note) by any amount obeys this inverse proportionality.



    Lets graph these tones using this website.
    440vs220 no overlay



    Overlaying them gives 44vs220 overlay



    Which results in
    enter image description here



    You can see from the first image how the waves change when you change octaves. From the second and third images, you can how the waves constructively and destructively interfere to form the final product that you hear. As you can see, the final product isn't a nice sine wave like the original two were, which makes me think that beats still exist but may be undetectable by ear if the two original notes are tuned well (this is just my suspicion, it's probably worth looking into yourself).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This sounds unlikely to me. For a wave like $sin t+sin[(2+epsilon)t]$, the mean squared amplitude doesn't vary slowly as it would in the case of beats, so I really doubt you're going to experience beats. Real musical sounds are not sine waves, so we expect beats between the harmonics that are close. This is well established in psychoacoustics.
      $endgroup$
      – Ben Crowell
      1 hour ago






    • 1




      $begingroup$
      When you superimpose two sine waves of different frequencies in physics, the phases will also matter for the resulting wave form. What you show, is therefore just one specific example. However, our hearing is insensitive to the phases. All superpositions of frequency $f$ and $2f$ differing in the phases will sound the same to us. I therefore think that you did not really answer the OP's question.
      $endgroup$
      – flaudemus
      56 mins ago










    • $begingroup$
      The note A2 is at 110 Hz. You mean A3 at 220 hertz. But anyway, this is not what beats are.
      $endgroup$
      – Pieter
      6 mins ago













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The phenomenon you're asking about is known as octave equivalence. It's a hard-wired thing in the human ear-brain system. (We know it's hard-wired because it's present without musical training and is true cross-culturally.) Notes that differ in frequency by a factor of 2 (or a power of 2) are perceptually similar, and may be mistaken for one another, even by trained musicians. So this is really a fact about the ear-brain system (i.e., psychology and physiology) rather than physics, although it does relate to a physical property.



    Looking through the index entries on this topic in Diana Deutsch, The psychology of music, 3rd ed., 2013, I don't see anything about any physical, neurological, or evolutionary explanation of octave equivalence. That may mean that nobody has such an explanation, or just that it wasn't something that Deutsch wanted to get into in this anthology of articles. It seems likely that it has at least some physical basis, because periodic tones usually have their first two frequencies (fundamental and first harmonic) in the ratio of two to one.




    An octave above a note is the fourth overtone, I'm assuming




    No, it's the first overtone.




    How do the sound waves compare between different octaves?




    Pitch perception can be complicated in some cases, but essentially our sense of pitch is normally based on the period of the wave. So notes differing by an octave in pitch have waveforms that are related by the fact that their periods are in a 2 to 1 ratio.




    You can still tune to these notes between octaves, and the beats can still be heard if out of tune.




    The beats would be beats between the first harmonic of the lower note and the fundamental of the higher note.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If you think about it, the human ear contains thousands (if not millions) of tiny hairs which vibrate at different frequencies. The nervous system detects the vibration as "sound". But, as with guitar strings, a hair that is "tuned" to a given frequency will tend to vibrate at frequencies twice as high, or, to a lesser extent, half as high. To avoid befuddling the poor human, the nervous system somehow combines the various related harmonics into a single sensation, with high or low being a separate sensation.
      $endgroup$
      – Hot Licks
      15 mins ago










    • $begingroup$
      @HotLicks: Normally a resonator does not respond strongly to a driving force at twice its resonant frequency, so it's far from obvious that the mechanism you propose should occur.
      $endgroup$
      – Ben Crowell
      8 mins ago










    • $begingroup$
      You're assuming linearity.
      $endgroup$
      – Hot Licks
      6 mins ago
















    3












    $begingroup$

    The phenomenon you're asking about is known as octave equivalence. It's a hard-wired thing in the human ear-brain system. (We know it's hard-wired because it's present without musical training and is true cross-culturally.) Notes that differ in frequency by a factor of 2 (or a power of 2) are perceptually similar, and may be mistaken for one another, even by trained musicians. So this is really a fact about the ear-brain system (i.e., psychology and physiology) rather than physics, although it does relate to a physical property.



    Looking through the index entries on this topic in Diana Deutsch, The psychology of music, 3rd ed., 2013, I don't see anything about any physical, neurological, or evolutionary explanation of octave equivalence. That may mean that nobody has such an explanation, or just that it wasn't something that Deutsch wanted to get into in this anthology of articles. It seems likely that it has at least some physical basis, because periodic tones usually have their first two frequencies (fundamental and first harmonic) in the ratio of two to one.




    An octave above a note is the fourth overtone, I'm assuming




    No, it's the first overtone.




    How do the sound waves compare between different octaves?




    Pitch perception can be complicated in some cases, but essentially our sense of pitch is normally based on the period of the wave. So notes differing by an octave in pitch have waveforms that are related by the fact that their periods are in a 2 to 1 ratio.




    You can still tune to these notes between octaves, and the beats can still be heard if out of tune.




    The beats would be beats between the first harmonic of the lower note and the fundamental of the higher note.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If you think about it, the human ear contains thousands (if not millions) of tiny hairs which vibrate at different frequencies. The nervous system detects the vibration as "sound". But, as with guitar strings, a hair that is "tuned" to a given frequency will tend to vibrate at frequencies twice as high, or, to a lesser extent, half as high. To avoid befuddling the poor human, the nervous system somehow combines the various related harmonics into a single sensation, with high or low being a separate sensation.
      $endgroup$
      – Hot Licks
      15 mins ago










    • $begingroup$
      @HotLicks: Normally a resonator does not respond strongly to a driving force at twice its resonant frequency, so it's far from obvious that the mechanism you propose should occur.
      $endgroup$
      – Ben Crowell
      8 mins ago










    • $begingroup$
      You're assuming linearity.
      $endgroup$
      – Hot Licks
      6 mins ago














    3












    3








    3





    $begingroup$

    The phenomenon you're asking about is known as octave equivalence. It's a hard-wired thing in the human ear-brain system. (We know it's hard-wired because it's present without musical training and is true cross-culturally.) Notes that differ in frequency by a factor of 2 (or a power of 2) are perceptually similar, and may be mistaken for one another, even by trained musicians. So this is really a fact about the ear-brain system (i.e., psychology and physiology) rather than physics, although it does relate to a physical property.



    Looking through the index entries on this topic in Diana Deutsch, The psychology of music, 3rd ed., 2013, I don't see anything about any physical, neurological, or evolutionary explanation of octave equivalence. That may mean that nobody has such an explanation, or just that it wasn't something that Deutsch wanted to get into in this anthology of articles. It seems likely that it has at least some physical basis, because periodic tones usually have their first two frequencies (fundamental and first harmonic) in the ratio of two to one.




    An octave above a note is the fourth overtone, I'm assuming




    No, it's the first overtone.




    How do the sound waves compare between different octaves?




    Pitch perception can be complicated in some cases, but essentially our sense of pitch is normally based on the period of the wave. So notes differing by an octave in pitch have waveforms that are related by the fact that their periods are in a 2 to 1 ratio.




    You can still tune to these notes between octaves, and the beats can still be heard if out of tune.




    The beats would be beats between the first harmonic of the lower note and the fundamental of the higher note.






    share|cite|improve this answer











    $endgroup$



    The phenomenon you're asking about is known as octave equivalence. It's a hard-wired thing in the human ear-brain system. (We know it's hard-wired because it's present without musical training and is true cross-culturally.) Notes that differ in frequency by a factor of 2 (or a power of 2) are perceptually similar, and may be mistaken for one another, even by trained musicians. So this is really a fact about the ear-brain system (i.e., psychology and physiology) rather than physics, although it does relate to a physical property.



    Looking through the index entries on this topic in Diana Deutsch, The psychology of music, 3rd ed., 2013, I don't see anything about any physical, neurological, or evolutionary explanation of octave equivalence. That may mean that nobody has such an explanation, or just that it wasn't something that Deutsch wanted to get into in this anthology of articles. It seems likely that it has at least some physical basis, because periodic tones usually have their first two frequencies (fundamental and first harmonic) in the ratio of two to one.




    An octave above a note is the fourth overtone, I'm assuming




    No, it's the first overtone.




    How do the sound waves compare between different octaves?




    Pitch perception can be complicated in some cases, but essentially our sense of pitch is normally based on the period of the wave. So notes differing by an octave in pitch have waveforms that are related by the fact that their periods are in a 2 to 1 ratio.




    You can still tune to these notes between octaves, and the beats can still be heard if out of tune.




    The beats would be beats between the first harmonic of the lower note and the fundamental of the higher note.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 3 hours ago

























    answered 3 hours ago









    Ben CrowellBen Crowell

    53k6164310




    53k6164310












    • $begingroup$
      If you think about it, the human ear contains thousands (if not millions) of tiny hairs which vibrate at different frequencies. The nervous system detects the vibration as "sound". But, as with guitar strings, a hair that is "tuned" to a given frequency will tend to vibrate at frequencies twice as high, or, to a lesser extent, half as high. To avoid befuddling the poor human, the nervous system somehow combines the various related harmonics into a single sensation, with high or low being a separate sensation.
      $endgroup$
      – Hot Licks
      15 mins ago










    • $begingroup$
      @HotLicks: Normally a resonator does not respond strongly to a driving force at twice its resonant frequency, so it's far from obvious that the mechanism you propose should occur.
      $endgroup$
      – Ben Crowell
      8 mins ago










    • $begingroup$
      You're assuming linearity.
      $endgroup$
      – Hot Licks
      6 mins ago


















    • $begingroup$
      If you think about it, the human ear contains thousands (if not millions) of tiny hairs which vibrate at different frequencies. The nervous system detects the vibration as "sound". But, as with guitar strings, a hair that is "tuned" to a given frequency will tend to vibrate at frequencies twice as high, or, to a lesser extent, half as high. To avoid befuddling the poor human, the nervous system somehow combines the various related harmonics into a single sensation, with high or low being a separate sensation.
      $endgroup$
      – Hot Licks
      15 mins ago










    • $begingroup$
      @HotLicks: Normally a resonator does not respond strongly to a driving force at twice its resonant frequency, so it's far from obvious that the mechanism you propose should occur.
      $endgroup$
      – Ben Crowell
      8 mins ago










    • $begingroup$
      You're assuming linearity.
      $endgroup$
      – Hot Licks
      6 mins ago
















    $begingroup$
    If you think about it, the human ear contains thousands (if not millions) of tiny hairs which vibrate at different frequencies. The nervous system detects the vibration as "sound". But, as with guitar strings, a hair that is "tuned" to a given frequency will tend to vibrate at frequencies twice as high, or, to a lesser extent, half as high. To avoid befuddling the poor human, the nervous system somehow combines the various related harmonics into a single sensation, with high or low being a separate sensation.
    $endgroup$
    – Hot Licks
    15 mins ago




    $begingroup$
    If you think about it, the human ear contains thousands (if not millions) of tiny hairs which vibrate at different frequencies. The nervous system detects the vibration as "sound". But, as with guitar strings, a hair that is "tuned" to a given frequency will tend to vibrate at frequencies twice as high, or, to a lesser extent, half as high. To avoid befuddling the poor human, the nervous system somehow combines the various related harmonics into a single sensation, with high or low being a separate sensation.
    $endgroup$
    – Hot Licks
    15 mins ago












    $begingroup$
    @HotLicks: Normally a resonator does not respond strongly to a driving force at twice its resonant frequency, so it's far from obvious that the mechanism you propose should occur.
    $endgroup$
    – Ben Crowell
    8 mins ago




    $begingroup$
    @HotLicks: Normally a resonator does not respond strongly to a driving force at twice its resonant frequency, so it's far from obvious that the mechanism you propose should occur.
    $endgroup$
    – Ben Crowell
    8 mins ago












    $begingroup$
    You're assuming linearity.
    $endgroup$
    – Hot Licks
    6 mins ago




    $begingroup$
    You're assuming linearity.
    $endgroup$
    – Hot Licks
    6 mins ago











    2












    $begingroup$

    I basically agree with the answers of Ben Crowell and of Pieter. The fact that we hear the octave as a pleasant pure interval is hard-wired in the human hearing system. As Pieter explained, musical instruments produce tones with many overtones of different intensities, where usually the frequency assigned to the tone is the lowest, and the overtones are integer multiples of it.



    The point I can add to these previous answers gives a hint, why the octave is a very special interval sounding most pure. The point is that if you produce a perfect octave on a musical instrument, the frequencies of all the overtones of the higher note will exactly line up with frequencies of the overtones of the lower note. This means that any beating in the resulting sound is reduced to a minimum.



    The situation is already different for a perfect fifth. In this case only every second overtone of the higher note lines up with an overtone of the lower note.



    The situation gets even worse for the other intervals. For example, for the minor third, only every fifth overtone lines up, and other overtones start to generate beatings. For dissonances, like the minor second, none of the overtones line up, and severe beatings arise from the overtones, resulting in a 'rough' perceived sound.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      I basically agree with the answers of Ben Crowell and of Pieter. The fact that we hear the octave as a pleasant pure interval is hard-wired in the human hearing system. As Pieter explained, musical instruments produce tones with many overtones of different intensities, where usually the frequency assigned to the tone is the lowest, and the overtones are integer multiples of it.



      The point I can add to these previous answers gives a hint, why the octave is a very special interval sounding most pure. The point is that if you produce a perfect octave on a musical instrument, the frequencies of all the overtones of the higher note will exactly line up with frequencies of the overtones of the lower note. This means that any beating in the resulting sound is reduced to a minimum.



      The situation is already different for a perfect fifth. In this case only every second overtone of the higher note lines up with an overtone of the lower note.



      The situation gets even worse for the other intervals. For example, for the minor third, only every fifth overtone lines up, and other overtones start to generate beatings. For dissonances, like the minor second, none of the overtones line up, and severe beatings arise from the overtones, resulting in a 'rough' perceived sound.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        I basically agree with the answers of Ben Crowell and of Pieter. The fact that we hear the octave as a pleasant pure interval is hard-wired in the human hearing system. As Pieter explained, musical instruments produce tones with many overtones of different intensities, where usually the frequency assigned to the tone is the lowest, and the overtones are integer multiples of it.



        The point I can add to these previous answers gives a hint, why the octave is a very special interval sounding most pure. The point is that if you produce a perfect octave on a musical instrument, the frequencies of all the overtones of the higher note will exactly line up with frequencies of the overtones of the lower note. This means that any beating in the resulting sound is reduced to a minimum.



        The situation is already different for a perfect fifth. In this case only every second overtone of the higher note lines up with an overtone of the lower note.



        The situation gets even worse for the other intervals. For example, for the minor third, only every fifth overtone lines up, and other overtones start to generate beatings. For dissonances, like the minor second, none of the overtones line up, and severe beatings arise from the overtones, resulting in a 'rough' perceived sound.






        share|cite|improve this answer











        $endgroup$



        I basically agree with the answers of Ben Crowell and of Pieter. The fact that we hear the octave as a pleasant pure interval is hard-wired in the human hearing system. As Pieter explained, musical instruments produce tones with many overtones of different intensities, where usually the frequency assigned to the tone is the lowest, and the overtones are integer multiples of it.



        The point I can add to these previous answers gives a hint, why the octave is a very special interval sounding most pure. The point is that if you produce a perfect octave on a musical instrument, the frequencies of all the overtones of the higher note will exactly line up with frequencies of the overtones of the lower note. This means that any beating in the resulting sound is reduced to a minimum.



        The situation is already different for a perfect fifth. In this case only every second overtone of the higher note lines up with an overtone of the lower note.



        The situation gets even worse for the other intervals. For example, for the minor third, only every fifth overtone lines up, and other overtones start to generate beatings. For dissonances, like the minor second, none of the overtones line up, and severe beatings arise from the overtones, resulting in a 'rough' perceived sound.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 23 mins ago

























        answered 29 mins ago









        flaudemusflaudemus

        1,853313




        1,853313























            1












            $begingroup$

            The human voice and most musical instruments produce tones with many overtones. The sound pressure is periodic with a period $T$, but does not vary like a sine wave at all. Instead, it can be described by a Fourier series. That is the sum of sines that are integer multiples of the fundamental frequency $f=1/T$.



            So when a singer sings with a fundamental at 220 hertz (an a), that note includes overtones at 440 hertz (a'), 660 hertz (e"), 880 hertz (a") etcetera. This is why beats can be heard between two instruments that play an a and an a' slightly out of tune.




            would the slopes of the waves align?




            The ear does a kind of Fourier analysis of the input waveform, where different frequencies are mapped to different parts of the cochlea and to different nerves going to the brain. Our hearing is almost insensitive to the relative phase between the frequency components.



            One can look at the frequency content of sounds with spectrogram apps on mobile phones or on https://musiclab.chromeexperiments.com/






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              This doesn't answer the question, which is about octave identification.
              $endgroup$
              – Ben Crowell
              3 hours ago










            • $begingroup$
              @BenCrowell The OP also wrote something about beats.
              $endgroup$
              – Pieter
              3 hours ago
















            1












            $begingroup$

            The human voice and most musical instruments produce tones with many overtones. The sound pressure is periodic with a period $T$, but does not vary like a sine wave at all. Instead, it can be described by a Fourier series. That is the sum of sines that are integer multiples of the fundamental frequency $f=1/T$.



            So when a singer sings with a fundamental at 220 hertz (an a), that note includes overtones at 440 hertz (a'), 660 hertz (e"), 880 hertz (a") etcetera. This is why beats can be heard between two instruments that play an a and an a' slightly out of tune.




            would the slopes of the waves align?




            The ear does a kind of Fourier analysis of the input waveform, where different frequencies are mapped to different parts of the cochlea and to different nerves going to the brain. Our hearing is almost insensitive to the relative phase between the frequency components.



            One can look at the frequency content of sounds with spectrogram apps on mobile phones or on https://musiclab.chromeexperiments.com/






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              This doesn't answer the question, which is about octave identification.
              $endgroup$
              – Ben Crowell
              3 hours ago










            • $begingroup$
              @BenCrowell The OP also wrote something about beats.
              $endgroup$
              – Pieter
              3 hours ago














            1












            1








            1





            $begingroup$

            The human voice and most musical instruments produce tones with many overtones. The sound pressure is periodic with a period $T$, but does not vary like a sine wave at all. Instead, it can be described by a Fourier series. That is the sum of sines that are integer multiples of the fundamental frequency $f=1/T$.



            So when a singer sings with a fundamental at 220 hertz (an a), that note includes overtones at 440 hertz (a'), 660 hertz (e"), 880 hertz (a") etcetera. This is why beats can be heard between two instruments that play an a and an a' slightly out of tune.




            would the slopes of the waves align?




            The ear does a kind of Fourier analysis of the input waveform, where different frequencies are mapped to different parts of the cochlea and to different nerves going to the brain. Our hearing is almost insensitive to the relative phase between the frequency components.



            One can look at the frequency content of sounds with spectrogram apps on mobile phones or on https://musiclab.chromeexperiments.com/






            share|cite|improve this answer











            $endgroup$



            The human voice and most musical instruments produce tones with many overtones. The sound pressure is periodic with a period $T$, but does not vary like a sine wave at all. Instead, it can be described by a Fourier series. That is the sum of sines that are integer multiples of the fundamental frequency $f=1/T$.



            So when a singer sings with a fundamental at 220 hertz (an a), that note includes overtones at 440 hertz (a'), 660 hertz (e"), 880 hertz (a") etcetera. This is why beats can be heard between two instruments that play an a and an a' slightly out of tune.




            would the slopes of the waves align?




            The ear does a kind of Fourier analysis of the input waveform, where different frequencies are mapped to different parts of the cochlea and to different nerves going to the brain. Our hearing is almost insensitive to the relative phase between the frequency components.



            One can look at the frequency content of sounds with spectrogram apps on mobile phones or on https://musiclab.chromeexperiments.com/







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 14 mins ago









            flaudemus

            1,853313




            1,853313










            answered 4 hours ago









            PieterPieter

            9,03331536




            9,03331536












            • $begingroup$
              This doesn't answer the question, which is about octave identification.
              $endgroup$
              – Ben Crowell
              3 hours ago










            • $begingroup$
              @BenCrowell The OP also wrote something about beats.
              $endgroup$
              – Pieter
              3 hours ago


















            • $begingroup$
              This doesn't answer the question, which is about octave identification.
              $endgroup$
              – Ben Crowell
              3 hours ago










            • $begingroup$
              @BenCrowell The OP also wrote something about beats.
              $endgroup$
              – Pieter
              3 hours ago
















            $begingroup$
            This doesn't answer the question, which is about octave identification.
            $endgroup$
            – Ben Crowell
            3 hours ago




            $begingroup$
            This doesn't answer the question, which is about octave identification.
            $endgroup$
            – Ben Crowell
            3 hours ago












            $begingroup$
            @BenCrowell The OP also wrote something about beats.
            $endgroup$
            – Pieter
            3 hours ago




            $begingroup$
            @BenCrowell The OP also wrote something about beats.
            $endgroup$
            – Pieter
            3 hours ago











            0












            $begingroup$

            The note $A4$ is $440Hz$. The note $A2$ is $220Hz$. They are one octave apart. We know that $s=lambda f$ where $s$ is the speed of sound in air, $lambda$ is wavelength, and $f$ is wave frequency. In terms of wavelength we have $$lambda=frac{s}{f}$$ So when we double the frequency going from $A2$ to $A4$, the wavelength is cut in half. Changing the frequency (note) by any amount obeys this inverse proportionality.



            Lets graph these tones using this website.
            440vs220 no overlay



            Overlaying them gives 44vs220 overlay



            Which results in
            enter image description here



            You can see from the first image how the waves change when you change octaves. From the second and third images, you can how the waves constructively and destructively interfere to form the final product that you hear. As you can see, the final product isn't a nice sine wave like the original two were, which makes me think that beats still exist but may be undetectable by ear if the two original notes are tuned well (this is just my suspicion, it's probably worth looking into yourself).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This sounds unlikely to me. For a wave like $sin t+sin[(2+epsilon)t]$, the mean squared amplitude doesn't vary slowly as it would in the case of beats, so I really doubt you're going to experience beats. Real musical sounds are not sine waves, so we expect beats between the harmonics that are close. This is well established in psychoacoustics.
              $endgroup$
              – Ben Crowell
              1 hour ago






            • 1




              $begingroup$
              When you superimpose two sine waves of different frequencies in physics, the phases will also matter for the resulting wave form. What you show, is therefore just one specific example. However, our hearing is insensitive to the phases. All superpositions of frequency $f$ and $2f$ differing in the phases will sound the same to us. I therefore think that you did not really answer the OP's question.
              $endgroup$
              – flaudemus
              56 mins ago










            • $begingroup$
              The note A2 is at 110 Hz. You mean A3 at 220 hertz. But anyway, this is not what beats are.
              $endgroup$
              – Pieter
              6 mins ago


















            0












            $begingroup$

            The note $A4$ is $440Hz$. The note $A2$ is $220Hz$. They are one octave apart. We know that $s=lambda f$ where $s$ is the speed of sound in air, $lambda$ is wavelength, and $f$ is wave frequency. In terms of wavelength we have $$lambda=frac{s}{f}$$ So when we double the frequency going from $A2$ to $A4$, the wavelength is cut in half. Changing the frequency (note) by any amount obeys this inverse proportionality.



            Lets graph these tones using this website.
            440vs220 no overlay



            Overlaying them gives 44vs220 overlay



            Which results in
            enter image description here



            You can see from the first image how the waves change when you change octaves. From the second and third images, you can how the waves constructively and destructively interfere to form the final product that you hear. As you can see, the final product isn't a nice sine wave like the original two were, which makes me think that beats still exist but may be undetectable by ear if the two original notes are tuned well (this is just my suspicion, it's probably worth looking into yourself).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This sounds unlikely to me. For a wave like $sin t+sin[(2+epsilon)t]$, the mean squared amplitude doesn't vary slowly as it would in the case of beats, so I really doubt you're going to experience beats. Real musical sounds are not sine waves, so we expect beats between the harmonics that are close. This is well established in psychoacoustics.
              $endgroup$
              – Ben Crowell
              1 hour ago






            • 1




              $begingroup$
              When you superimpose two sine waves of different frequencies in physics, the phases will also matter for the resulting wave form. What you show, is therefore just one specific example. However, our hearing is insensitive to the phases. All superpositions of frequency $f$ and $2f$ differing in the phases will sound the same to us. I therefore think that you did not really answer the OP's question.
              $endgroup$
              – flaudemus
              56 mins ago










            • $begingroup$
              The note A2 is at 110 Hz. You mean A3 at 220 hertz. But anyway, this is not what beats are.
              $endgroup$
              – Pieter
              6 mins ago
















            0












            0








            0





            $begingroup$

            The note $A4$ is $440Hz$. The note $A2$ is $220Hz$. They are one octave apart. We know that $s=lambda f$ where $s$ is the speed of sound in air, $lambda$ is wavelength, and $f$ is wave frequency. In terms of wavelength we have $$lambda=frac{s}{f}$$ So when we double the frequency going from $A2$ to $A4$, the wavelength is cut in half. Changing the frequency (note) by any amount obeys this inverse proportionality.



            Lets graph these tones using this website.
            440vs220 no overlay



            Overlaying them gives 44vs220 overlay



            Which results in
            enter image description here



            You can see from the first image how the waves change when you change octaves. From the second and third images, you can how the waves constructively and destructively interfere to form the final product that you hear. As you can see, the final product isn't a nice sine wave like the original two were, which makes me think that beats still exist but may be undetectable by ear if the two original notes are tuned well (this is just my suspicion, it's probably worth looking into yourself).






            share|cite|improve this answer









            $endgroup$



            The note $A4$ is $440Hz$. The note $A2$ is $220Hz$. They are one octave apart. We know that $s=lambda f$ where $s$ is the speed of sound in air, $lambda$ is wavelength, and $f$ is wave frequency. In terms of wavelength we have $$lambda=frac{s}{f}$$ So when we double the frequency going from $A2$ to $A4$, the wavelength is cut in half. Changing the frequency (note) by any amount obeys this inverse proportionality.



            Lets graph these tones using this website.
            440vs220 no overlay



            Overlaying them gives 44vs220 overlay



            Which results in
            enter image description here



            You can see from the first image how the waves change when you change octaves. From the second and third images, you can how the waves constructively and destructively interfere to form the final product that you hear. As you can see, the final product isn't a nice sine wave like the original two were, which makes me think that beats still exist but may be undetectable by ear if the two original notes are tuned well (this is just my suspicion, it's probably worth looking into yourself).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            roshokaroshoka

            719




            719












            • $begingroup$
              This sounds unlikely to me. For a wave like $sin t+sin[(2+epsilon)t]$, the mean squared amplitude doesn't vary slowly as it would in the case of beats, so I really doubt you're going to experience beats. Real musical sounds are not sine waves, so we expect beats between the harmonics that are close. This is well established in psychoacoustics.
              $endgroup$
              – Ben Crowell
              1 hour ago






            • 1




              $begingroup$
              When you superimpose two sine waves of different frequencies in physics, the phases will also matter for the resulting wave form. What you show, is therefore just one specific example. However, our hearing is insensitive to the phases. All superpositions of frequency $f$ and $2f$ differing in the phases will sound the same to us. I therefore think that you did not really answer the OP's question.
              $endgroup$
              – flaudemus
              56 mins ago










            • $begingroup$
              The note A2 is at 110 Hz. You mean A3 at 220 hertz. But anyway, this is not what beats are.
              $endgroup$
              – Pieter
              6 mins ago




















            • $begingroup$
              This sounds unlikely to me. For a wave like $sin t+sin[(2+epsilon)t]$, the mean squared amplitude doesn't vary slowly as it would in the case of beats, so I really doubt you're going to experience beats. Real musical sounds are not sine waves, so we expect beats between the harmonics that are close. This is well established in psychoacoustics.
              $endgroup$
              – Ben Crowell
              1 hour ago






            • 1




              $begingroup$
              When you superimpose two sine waves of different frequencies in physics, the phases will also matter for the resulting wave form. What you show, is therefore just one specific example. However, our hearing is insensitive to the phases. All superpositions of frequency $f$ and $2f$ differing in the phases will sound the same to us. I therefore think that you did not really answer the OP's question.
              $endgroup$
              – flaudemus
              56 mins ago










            • $begingroup$
              The note A2 is at 110 Hz. You mean A3 at 220 hertz. But anyway, this is not what beats are.
              $endgroup$
              – Pieter
              6 mins ago


















            $begingroup$
            This sounds unlikely to me. For a wave like $sin t+sin[(2+epsilon)t]$, the mean squared amplitude doesn't vary slowly as it would in the case of beats, so I really doubt you're going to experience beats. Real musical sounds are not sine waves, so we expect beats between the harmonics that are close. This is well established in psychoacoustics.
            $endgroup$
            – Ben Crowell
            1 hour ago




            $begingroup$
            This sounds unlikely to me. For a wave like $sin t+sin[(2+epsilon)t]$, the mean squared amplitude doesn't vary slowly as it would in the case of beats, so I really doubt you're going to experience beats. Real musical sounds are not sine waves, so we expect beats between the harmonics that are close. This is well established in psychoacoustics.
            $endgroup$
            – Ben Crowell
            1 hour ago




            1




            1




            $begingroup$
            When you superimpose two sine waves of different frequencies in physics, the phases will also matter for the resulting wave form. What you show, is therefore just one specific example. However, our hearing is insensitive to the phases. All superpositions of frequency $f$ and $2f$ differing in the phases will sound the same to us. I therefore think that you did not really answer the OP's question.
            $endgroup$
            – flaudemus
            56 mins ago




            $begingroup$
            When you superimpose two sine waves of different frequencies in physics, the phases will also matter for the resulting wave form. What you show, is therefore just one specific example. However, our hearing is insensitive to the phases. All superpositions of frequency $f$ and $2f$ differing in the phases will sound the same to us. I therefore think that you did not really answer the OP's question.
            $endgroup$
            – flaudemus
            56 mins ago












            $begingroup$
            The note A2 is at 110 Hz. You mean A3 at 220 hertz. But anyway, this is not what beats are.
            $endgroup$
            – Pieter
            6 mins ago






            $begingroup$
            The note A2 is at 110 Hz. You mean A3 at 220 hertz. But anyway, this is not what beats are.
            $endgroup$
            – Pieter
            6 mins ago












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