3
$\begingroup$

All the research I've been doing tells me that a guitar note is determined by the fundamental frequency played. But say you play an A on the open A string (110 Hz), and then play a higher octave A by pressing behind a certain fret on another string (220 Hz). Why is this a different A, when it is just a harmonic of the fundamental frequency? Does this mean that the fundamental frequency is now the 220 Hz, and the first harmonic of 110 Hz is no longer played? Is this correct?

$\endgroup$
  • 2
    $\begingroup$ I'm voting to close this question as off-topic because it is not about physics. $\endgroup$ – Emilio Pisanty Dec 24 '19 at 21:36
  • $\begingroup$ @EmilioPisanty Since when Fourier analysis of vibrating strings and sound waves is not physics? $\endgroup$ – GiorgioP Dec 25 '19 at 19:37
  • $\begingroup$ @GiorgioP As phrased the question is about music playing, not Fourier analysis. $\endgroup$ – ZeroTheHero Dec 26 '19 at 2:21
  • 1
    $\begingroup$ @ZeroTheHero The question, as it has been phrased, is clearly about the presence or not of Fourier components in the sound coming from a string. Mentioning a guitar is just putting the question in its context which is applied physics. I have studied music as well as physics and I know the difference. Moreover, as you can see, the answers are about the physical presence of Fourier components, although in the language of musical acustics they are called overtones. $\endgroup$ – GiorgioP Dec 26 '19 at 8:07
  • $\begingroup$ @EmilioPisanty This question is clearly a physics question relating the length of a string to it's fundamental frequency and harmonics. $\endgroup$ – BioPhysicist Dec 26 '19 at 11:59
2
$\begingroup$

When you push down on the fret you are essentially changing the length of the string. This means you are chaining the fundamental frequency.

This means that, while the $220\ \rm{Hz}$ is a harmonic of the string with the $110\ \rm{Hz}$ fundamental frequency, when you half the length of the string the fundamental frequency is now $220\ \rm{Hz}$, and the $110\ \rm{Hz}$ is no longer a harmonic of the shortened string.

Mathematically, the modes of the string are achieved when the string of length $L$ is broken up into parts such that $$\lambda_n=\frac{2L}{n}$$ Where $\lambda_n$ is the wavelength of the standing wave, and $n$ is a positive integer. Since $v=f\lambda$ is true for the waves, where $v$ is the wave velocity that depends on the string properties, we have $$f_n=\frac{nv}{2L}$$

Since the fundamental frequency is when $n=1$, if $f_1=220\ \rm{Hz}$, then this is the lowest $f_n$ can be for larger $n$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The question you ask is less physical and more psychological. You are correct, that there's nothing different between a 220Hz fundamental and 220Hz 1st harmonic of a 110Hz from a wave perspective. Both behave the same.

It's how the ear processes it that is different. Our brains are built with an awareness of these harmonics, and it's effective to sort of bundle the fundamental and the harmonics together into a "pitch," driven primarially by the fundamental, and the "color" of the note, which is related to the ratios of various harmonics. Our brain does this process naturally, so we often don't even notice the effects.

In your case, if you played an A110 and an A220, you would get a set of harmonics. A110 would create harmonics at 220, 330, 440, etc. A220 would create harmonics at 440, 880, etc. The brain would notice the presence of those 110Hz, 330Hz, and 550Hz signals all belong as part of one "note," processing it as an "A" (describing the pitch), with a particular color. It would then be left with a set of frequencies which fit well with an "A one octave up", with a similar color.

In effect, your brain will divide up the amplitude of the 220Hz, 440Hz, 660Hz, etc. components of the sound between the expected harmonics of a string vibrating at 110Hz, and what is left over will be treated as a second note, whose fundamental is 220Hz.

This effect can be messed with. Throat singers are famous for doing this. They sing a very low note, then alter the shape of their mouth to cause one of the overtones to be much more resonant. When our ears hear this, we get the impression that that overtone isn't part of the usual harmonic series of a human voice because its so much louder than the harmonics nearby. Thus, what we hear is something akin to someone singing two notes at the same time.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It might be worth noting that you can also fool the ear into inventing a fundamental which is not there. If you have a sound which looks like $\sum_{n=2}\alpha_n \sin(n\omega t)$ (so with the fundamental missing), you will often hear this as a note an octave down. $\endgroup$ – tfb Oct 9 '18 at 6:50
1
$\begingroup$

In addition to Aaron's and Cort's answers, you can prove that the 220 Hz first overtone is actually present when you pluck the open A (110 Hz) by merely touching the string at the 12th fret, which is the midpoint of the string. This dampens the fundamental component because it has an antinode at that location, but it leaves the 1st overtone/2nd harmonic ringing because it has a node in the middle. The 3rd overtone/4th harmonic is also ringing but the 2nd overtone is gone.

You can also touch the string at the 12th fret and pluck it to get a different sound because both halves of the string will be vibrating out of phase.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Your two tones have two different spectra:

open string     | octave (fretted)
----------------+-----------------
A    (110Hz)    |
a    (220Hz)    | a    (220Hz)
e'   (330Hz)    |
a'   (440Hz)    | a'   (440Hz)
c#'' (550Hz)    |
e''  (660Hz)    | e''  (660Hz)
g''  (770Hz)    |
a''  (880Hz)    | a''  (880Hz)

You see, in the fretted case, only every other overtone of the open string is present. The fundamental of the open string and all its odd multiples are missing entirely. This is enough for your ear to distinguish the two sounds as two different notes.

However, as you correctly noticed, the octave and all of its overtones are found as overtones in the spectrum of the open string. This is why the octave blends so well into the sound of the lower note. If the octave is played simultaneously with the lower note, and not played too loud, your ear does not really recognize it as a note of its own, it just recognizes that the overtone spectrum of the lower note has an odd wavy shape, which changes the perceived sound.

This is a technique used frequently in piano music: The left hand does not just play the base note of the chord, but rather the base note plus its octave to give the note a mightier sound. The octave is not meant to be perceived as a note of its own, it's meant to blend into the base note, enhancing its sound.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.