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When you pluck a single guitar string, several different frequencies/tones are produced at once. leftaroundbout's answer here does a great job of explaining why these higher harmonics are produced on a guitar: when you pull back the string, you produce a shape that does not match a simple sinusoidal shape and thus doesn't match a single frequency/pitch. Rather, the complex shape can only be created from a combination of simpler sinusoidal waves/pure pitches. These simpler sinusoidal waves that combine together to form the complex shape are the harmonic frequencies. Producing the complex shape in turn produces those harmonic frequencies, which we then hear as higher pitches.

As an example, we could give the string a shape that consists of:

  • 50% of the fundamental frequency f1,

  • 30% of the second harmonic f2,

  • 15% of the third harmonic f3, and

  • 5% of the fourth harmonic f4.

    I'm making up these numbers. I imagine the true coefficients when plucking a string would be closer to a sawtooth waveform.

The percentages/weights (more formally called Fourier coefficients) determine how loud the harmonics are.

Presumably, there are other mechanisms which also determine how loud the harmonics are--mechanisms beyond just the starting shape of the string when plucked. I don't know what these other mechanisms are, but I'm imagining that the string vibrates slightly at its ends and interacts with the other parts of the guitar. I would expect that these sorts of interactions impact the strength (amplitude/intensity) of the harmonic frequencies. (For example, maybe certain frequencies are damped more strongly.)

My question isn't about the mechanisms--although folks are free to weigh in on those as well. Rather, here is my question: how much do the relative volumes of the harmonic frequencies vary when plucking a guitar string? Does plucking the guitar string at different locations along the string impact how loud certain harmonics are? I'm not talking about muting the fundamental--I'm referring to plucking a guitar string once and simply letting it ring.

3 Answers 3

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+50

It's easier to give a simple but physically correct answer to a different but related problem: "How much do the relative volumes of the harmonics vary when you hit a string, rather than pluck it." That corresponds exactly to how a piano or a dulcimer works, or course.

In that case, you can understand what happens by looking at the shape of the vibration for each harmonic. For each harmonic, the maximum amplitude looks like a sine-wave along the length of the string. The amplitude is zero at both ends and there are 0, 1, 2, ... "nodes" spaced evenly along the length (zero nodes for the fundamental, 1 for the second harmonic, etc) where the amplitude is zero.

The amount by which each harmonic is excited depends on the vibration amplitude at the point where you hit the string. For example if you hit it at the mid-point, you would not excite the second harmonic at all, because it has a node at the striking point. The same is true for all the even harmonics. So you would excite only the odd harmonics 1, 3, 5, etc.

In fact, pianos are designed following this principle. If the fundamental frequency is the note C, the harmonics follow the pattern C, C, G, C, E, G, "Bb", C, etc. The 7th harmonic labelled "Bb" is actually flatter than Bb in he standard Western (equal temperament) scale and therefore gives an out-of-tune quality to the timbre in polyphonic music. The piano hammer is designed to hit the string at 1/7 of the length of the string, where this harmonic has a node, and therefore (in an ideal world!) it is not excited at all and the tone of the instrument is improved.

Hitting the string at the 1/7 position also suppresses the 14th, 21st, 28th, etc, harmonics, but such high harmonics don't have a large influence on the tone, and they are damped out very quickly compared with the fundamental in any case.

For a plucked string, the general idea is the same, but the details are different. it is a reasonable approximation to assume that you hold the string "bent into a two straight segments" at the plucking point, and then let it go. To find how the make that deformed shape, by combining the harmonics, you have to do a mathematical process (Fourier analysis). The only case where the solution is "obvious" is if you pluck the string at the mid point, where you get the same result as for hitting the string - you don't excite any even harmonics at all.

As a general principle, if you pluck nearer the end of the string, you excite the high harmonics more compared with the low ones, without suppressing any of them. Thus the tone of an acoustic guitar sounds "brighter" when you pluck closer to the bridge - but if you get too close to the bridge, the string is harder to pluck so the amplitude of all the harmonics is less, and the volume of the note is quieter.

In a real instrument like a guitar, this becomes more complicated because of the resonances of the body of the instrument itself. That has a set of harmonics which are independent of the note being played, which don't have any simple ratio of frequencies like 1:2:3:..., and which are damped out quickly. But they produce an audible "thump" which obscures the start of the "real" note from the vibrating string and changes the listener's perception of how the tone of the note sounds. Of course you can excite the resonances of the body on their own, which is a basic technique in some styles of playing (e.g. flamenco) - just mute the strings and tap the guitar with your fingers!

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  • Assuming by second harmonic you mean an octave above the open string (the half-way point). Hitting the string at that point does produce that harmonic - called a 'tapped harmonic', even though it's being hit on a node. There's some confusion as to what to call the fundamental note - it's sometimes referred to as the first harmonic.
    – Tim
    Aug 5, 2017 at 16:03
  • @alephzero, when you strike a string at, say, the midway point, what determines the coefficients/relative strengths of the odd harmonics? Let's assume it's an ideal string being struck which is perfectly tied down at its ends (the two boundaries are true nodes). Is the 3rd harmonic excited only 1/3 the amount of the 1st harmonic because the 3rd harmonic is only being struck at one of three antinodes? Similarly, striking at the middle only activates the 5th harmonic at 1 of 5 antinodes, so is it 1/5 as strong as the first harmonic? Is this sort of divvying how the amplitudes are determined?
    – jdjazz
    Aug 6, 2017 at 3:15
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Yes, where you pluck has a major effect on the volume of these partials, so you can dramatically change the tone.

In fact you should try this yourself:

Pluck (or pick) really close to the bridge = you will notice a jangly, bright tone - if you want you can relate this back to the shape of that initial bend of the string as the pick or your fingers release it generating louder high harmonics.

Then compare with a pluck close to the 12th fret - much more mellow, more mids, and much less jangle. More jazzy, if you like. This is because you have a lower amplitude of those higher harmonics.

I have popped a brief example up on Youtube.

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Well, yes – as Tim said, and I did earlier in the linked post, plucking the string at different locations affects how loud various harmonics are in the way that the initial spectrum is essentially the Fourier transform of the triangular shape the string has when deformed through the plucking, before releasing it.

It's worth some explanation why this is so. Forgetting about frequency analysis for a moment – the string can very well be modelled by the wave equation

∂²ψt² = Fρ · ∂²ψx²

where ψ(x,t) is the displacement from rest of the string at position x and time t, and c² = Fρ is the (constant) ratio of string tension and mass density along the string. (The wave equation is basically just Newton's law of motion for an infinitesimally small bent piece of string under tension.)

It is easy to come up with solutions to this equation even if you don't know anything about trigonometry / Fourier series. Namely, starting out with any function ψ₀(x) ≡ ψ(x,0), you can define

ψ(x,t) := ψ₀(x + c·t)

and calculate

∂²ψt² = t (c · xψ₀(x + c·t))
            = c² · ∂²x²ψ₀(x + c·t)
            = c² · ∂²ψx²

I.e. this works for any starting function, you just do time evolution by moving the entire shape to the left with phase velocity c.

As a result, the temporal waveform (which is what's picked up by the bridge and thus transmitted to the air and eventually your ears) is essentially the same as the spatial waveform of the initial deformation of the string. In particular, if you pluck the string in the middle, you'll get pretty much a triangular wave which is known to sound soft/sweet with no even harmonics, and if you pluck very close to the bridge you have pretty much a sawtooth, which not only has the even overtones very present, it also has generally much less amplitude decay towards the high harmonics. The reason for this is that at the beginning, you have a very strong spatial derivative close to the bridge, almost a discontinuity. That gives essentially a sharp transient whenever the travelling wave packet hits the bridge.


The left-moving wave is not the only solution: just as valid is

ψ(x,t) := ψ₀(xc·t)

∂²ψt² = t (-c · xψ₀(x + c·t))
            = +c² · ∂²x²ψ₀(x + c·t)
            = c² · ∂²ψx².

Since the wave equation is a second order linear PDE, the general solution is obtained by combining a left-moving contribution with a right-moving contribution,

ψ(x,t) = A·ψ(x,t) + A·ψ(x,t)

with the proportion of amplitudes determined by what vertical velocity the start state has, because that velocity is determined by the difference of the amplitude factors:

ψt = A·c·ψ(x,t) − A·c·ψ(x,t)

i.e.

ψt|t=0 = (AAc·ψ₀(x).

In particular, for a start state with essentially no movement, you need A = A. Thus, the string movement will consist to equal proportions of a left-travelling wave and a right-travelling wave.

Even if you managed to start out with only the left-moving part: each time the wave hits the bridge, most of the signal get reflected back onto the string, travelling in opposite direction.

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