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A guitar is a plucked instrument and it is played by plucking a string at an off-centre point fixed at two ends. In general, Fourier analysis tells that all harmonics (the resonant frequencies of the string) will be excited and the string will vibrate in a superposition of different harmonics. It is true that for $n^{th}$ harmonic the amplitude goes like $1/n$ suggesting that the fundamental contributes the most. In this situation, I am confused. How does a guitar string produce a pure tone/pure frequency sound instead of a noise?

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  • $\begingroup$ Related: physics.stackexchange.com/q/429495 $\endgroup$
    – PM 2Ring
    Commented Mar 3, 2019 at 9:49
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    $\begingroup$ Note: It's also a myth that all harmonics are generated by plucking. Pluck a string exactly in the middle, and you get a tone that only includes the even harmonics. This tone sounds very different from the tone that you get by plucking near the end of the string where you get a near complete set of harmonics. Also, the relative strengths of the harmonics varies with plucking location, which also contributes to slight changes of sound. Try it out if you have a guitar handy! E-guitarists use this effect a lot in conjunction with pick-up location to enable sounds otherwise impossible. $\endgroup$ Commented Mar 3, 2019 at 17:07
  • $\begingroup$ You can reduce the number of harmonics by playing a string harmonic. It sounds quite different from the same pitch played "normally" which contains the full set of harmonics. $\endgroup$ Commented Mar 3, 2019 at 23:01
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    $\begingroup$ It doesn't. Aside from the accurate comments so far, a guitar is not a perfect resonator (you know, massless string with perfect nodes at the ends, and so on). In real life, then, the amplitude of the original pluck plays strongly into the overtone series magnitudes. $\endgroup$ Commented Mar 4, 2019 at 13:44
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    $\begingroup$ What should be brought into the discussion is how many string frequencies will vibrate the wooden body of the instrument. Is not that where the sound comes from? Electric guitars sounds are of interest also. Here, one may imagine filters so only certain pitches get amplified. Stradivarius may have known something about this (for violins). $\endgroup$ Commented Dec 8, 2022 at 1:10

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Usually, a guitar does not produce a pure tone/frequency. If so, its sound would be very close to a diapason. The difference between noise and a musical tone is not that a unique frequency makes a tone. Still, there is a continuum between a pure tone (one frequency) and noise (all frequencies, not only multiple of a fundamental, without any regular pattern among their weights), where many non-pure tones are still recognized as dominated by a fundamental frequency. The additional frequencies add what we call the tone color or timbre of the sound.

In general, each harmonic's exact weight can vary according to how and where the chord is plucked. You might find interesting this study on the subject.

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Human perception is involved here because when you humans talk about noise this generally means a sound that is aperiodic. However the tone produced by a guitar will be something like:

$$ A(t,x) = \sum_{i=0}^\infty A_i \sin(n\omega_i t - k_i x) $$

i.e. a superposition of the frequencies $f$, $2f$, $3f$, etc. The function $A(t,x)$ is periodic in time with frequency $f = 2\pi\omega_0$ so the ear/brain team perceives it as a tone not a noise.

Constructing a noise is actually quite complicated as we need to include all frequencies, not just integer multiples of a fundamental, and there will be a phase term in the equation that is not constant i.e. the sine waves making up the noise are not coherent.

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  • $\begingroup$ I think I may ask a related question here. What it means to tune a detuned guitar and why do different strings make different sounds? $\endgroup$ Commented Mar 3, 2019 at 10:06
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    $\begingroup$ @mithusengupta123 That's probably better asked on the Music SE. The details of what overtones are produced will be a complicated function of lots of factors like the string density and it's position wrt the body of the guitar. Predicting the frequency spectrum from first principles using physics is impossibly hard. I guess some form of numerical finite element calculation would be needed. $\endgroup$ Commented Mar 3, 2019 at 10:13
  • $\begingroup$ OK. I think that the only parameter than can change from one string to another is the tension T because the length L is same for all strings. That can change the fundamental and overtones from one string to another. Not very sure though $\endgroup$ Commented Mar 3, 2019 at 10:23
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    $\begingroup$ As indicated, it is a complicated subject. On a first approach, you can consider that adjusting the tension just fix the frequency of the fundamental: the pitch of the sound. The way you pluck the string, the fixation of the string to the guitar, the fixation to the soundboard .....fix the proportion of harmonics which is related to the timbre of the note. $\endgroup$ Commented Mar 3, 2019 at 14:12
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    $\begingroup$ "you humans"? Was this answer generated by a robot? $\endgroup$
    – Barmar
    Commented Mar 4, 2019 at 17:14
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For an ideal string, the key point is that all the harmonics are "harmonic" : their frequency is a integer multiple of the frequency of the fundamental. So the movement of the string is periodic and has a well definite frequency.

For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}\cos (2\pi {{f}_{1}}t)+{{A}_{2}}\cos (2\pi 2{{f}_{1}}t)+{{A}_{3}}\cos (2\pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$

To complete, the dependence in $n$ is rather $\approx 1/{{n}^{2}}$ for a plucked string.

Sorry for my english !

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  • $\begingroup$ Not clear. All harmonics are present at the same time and the string doesn't vibrate with a particular resonant frequency $\endgroup$ Commented Mar 3, 2019 at 9:11
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    $\begingroup$ For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}\cos (2\pi {{f}_{1}}t)+{{A}_{2}}\cos (2\pi 2{{f}_{1}}t)+{{A}_{3}}\cos (2\pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$ $\endgroup$ Commented Mar 3, 2019 at 9:14
  • $\begingroup$ OK. I have completed ! $\endgroup$ Commented Mar 3, 2019 at 9:27
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You do not hear a pure note. What is presented to your ear is a whole mixture of frequencies, which are approximately integer multiples of each other. It is your brain that melds these sounds together to give you wnat sounds like a single pitch (with a variety of "tone colours" depending on the mixture of frequencies.

To see the effect, take the top string of the guitar and wrap a small piece of masking tape around it right in the middle of the length of the string. (small means less that a square 1/2 inch on a side). You will now hear two notes, one approximately an octave higher than the other. Your brain now does not meld the even harmonics (frequencies which are even whole number multiples of the fundamental frequency of the string) and the odd, and instead hears them as two separate notes.

The frequencies that go into the ear are almost the same here as they were when you plucked the string without the tape, so the processing to produce one or two notes out of those frequencies takes place in the brain. Some argue that it is the periodicity of the tone, but it is clear that you hear one note even if someone in the room is talking while you play the note. Their talking ruins any periodicity in the sound coming into your ear. But still you hear only one note (or two if you put the tape on).

Another instance in which the ability of the brain to meld the harmonics into one note is modified is with Tuvan Singit throat singing. It sounds like a drone and a high pitch flute over the drone. But the high pitch is an exact multiple of the drone, and your brain, one would think, would meld the two into one note (as it does in normal singing).

How or why the brain does this is still, AFAIK, poorly understood.

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  • $\begingroup$ Do you think you could break this up into paragraphs? It's kinda hard to read now. $\endgroup$
    – pipe
    Commented Mar 4, 2019 at 9:17
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All of the frequencies that don't satisfy the boundary conditions will very quickly cancel with their own reflections off the boundaries due to phase relationships.

Within a split second after the string is plucked, any recording will show very clean harmonics in the Fourier spectrum that will all be integer multiples of the fundamental. There are lots of phone apps that'll do this these days, so try it out for yourself.

The relative amplitudes of the various harmonics create the timbre of the guitar, for instance, once the string is resounding. Bottom line is there is nothing like broad spectrum noise coming from a guitar string except for the split second after the string is plucked, and that is nowhere near the duration of a short note played in a quick piece of music.

Human perception certainly plays into music. We tend to perceive the lowest tone as the note being played, even though there are many tones coming from that one string. The other overtones play into the "character" or timbre of the instrument instead of being perceived as pitch. The relative strengths of the overtones depend on the details of the instrument - it's construction, geometry, etc. The air mass inside the guitar, for instance, needs to resound with the strings in order to amplify the sound in an acoustic guitar. It's complex shape is a marriage of needing to amplify lots of different wavelengths coming from the strings, while also being comfortable to hold... no easy engineering feat, but one which is easy to take for granted.

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