I understand that when you pluck a guitar string, then a bunch of harmonic frequencies are produced rather than just the frequency of the desired note.

If this is true, why does C2 sound so different from C1? I mean, C2 is a harmonic of C1, and should therefore be heard when C1 is played. Why are all these harmonics produced on top of the targeted note in the first place?

Also, what happens if you don't pluck at the center of a string?

  • $\begingroup$ Somewhat related question here on Math.SE. $\endgroup$
    – Nikolaj-K
    Jul 1, 2012 at 22:08

3 Answers 3


Let's look at frequency instead of notes. Let's say the string has a natural frequency of $100 Hz$ and that harmonics are present when you pluck it. Then, the frequency content of the sound will be of the form:

$a_1 \cdot 100 Hz + a_2 \cdot 200 Hz + a_3 \cdot 300 Hz + ... $

Now, let's say you fret this string halfway such that the natural frequency becomes $200 Hz$. When plucked, the frequency content of the sound will be of the form:

$b_1 \cdot 200 Hz + b_2 \cdot 400 Hz + b_3 \cdot 600 Hz + ... $

See the difference? The 2nd sound is missing many frequencies that the first sound contained.

There are a number of reasons harmonics are produced. One is that when you pluck the string, the initial configuration is not a pure sinusoid but more like a sawtooth or triangle. It's easy to show mathematically that a sawtooth or triangle shape can be "built up" from the fundamental and harmonics. Only the the sinusoid has a single frequency.

Plucking at a different point changes the initial configuration and thus the frequency content.

  • $\begingroup$ Which component has the highest amplitude? Does it depend on where the string is plucked? $\endgroup$
    – leongz
    Jul 1, 2012 at 23:55
  • 1
    $\begingroup$ Generally speaking, the fundamental has the highest amplitude. However, by lightly resting your finger at, for example, the halfway point, and light plucking the string while lifting your finger, you can kill the fundamental and the string will vibrate primarily at the 2nd harmonic. $\endgroup$ Jul 2, 2012 at 0:18
  • 2
    $\begingroup$ To build on what @AlfredCentauri said, an example of this is the high-sounding notes that you hear in, for instance, the intro riff to Pearl Jam's "Jeremy" $\endgroup$ Jul 2, 2012 at 3:17

Frequency is just a way of analyzing a time dependent motion. Consider plucking a string by first pulling one point on the string away from its equilibrium. The string shape will be like a triangle, two straight bits of string coming away from where your finger is holding the string, but meeting at a slight angle where your finger holds the string.

That triangle can be expressed as a sum of sinusoids through fourier analysis. We know the ends of the string are constrained to be at 0, so we know that only fourier components that have 0 at the two ends of the strings are used in the fourier expansion. So we have $$ S(x) = a_n sin( n\pi x / L )$$ where $L$ is the distance from where the string is attached at one end to where the string is attached at the other end, and we assign $x=0$ at one attachment point, so $x=L$ at the other, and $S(x)$ is the displacement of the string at all points $0<x<L$ in between.

Amazingly, each of these spatial Fourier components will correspond to a temporal frequency component, one of the harmonics, which we will see when we let the string go (finish our pluck). The string will not maintain its triangular shape because the different fourier components will evolve at different speeds.

So we have a complicated time dependent shape of the plucked string $S(x,t)$ which happnes to be expressible as the sum over sinusoidal shapes of the string that start out as the triangular shape of the initial pluck.

Connection to Time Domain

Above we wrote the spatial equations as a fourier series. A better poster than myself would actually find the values of $a_n$ to add up to a nice triangular wave for you, but I won't do that. But those are the values of $a_n$ you woul want.

But we can do better. Each spatial fourier component $sin(n\pi x/L)$ has a specific harmonic time evolution associated with it $cos(n\omega_0 t)$. $f_n = n w_0/(2\pi)$ are the harmonic frequencies previously talked about. So we actually have a time dependent solution for the motion of the released string: $$ S(x,t) = a_n sin( n\pi x / L )cos(n\omega_0 t).$$

If we could pluck the string initially into the shape of half a sine wave extending from x=0 to x=L, when it was let go, it would resonate ONLY at the fundamental frequency f_0. But The triangle shape of the string is decomposed into a variety of sine wave components, each one of which will time-evolve faster than the lower harmonics. The net evolution of $S(x,t)$ will be something beautiful to watch indeed, and if you guys started paying me I would write the matlab code to do the animation and figure out how to post it. But for free, you will have to be motivated to code this up yourself to see it. Suffice it to say, once the motion starts, the string does not look like a triangle wave any longer.

In summary, because our original pluck has a shape to the string which can be written as a Fourier series over many different sine wave components, the string when it moves will have harmonic motion at many different frequencies, all harmonics of the frequency of the $n=1$ longest sine wave time-variation.

We have not attempted to write the time- and space- differential equation from which this solution comes. You will learn this all in due course. We have simply asserted a solution that at least makes intuitive sense: the higher the "spatial frequency" of the sine wave component of the string, the higher the temporal frequency associated with that spatial component, and this is how we get all those harmonics in our plucked string.

  • $\begingroup$ what do you mean "the different fourier components will evolve at different speeds?" $\endgroup$
    – Griffin
    Jul 2, 2012 at 3:27
  • $\begingroup$ @Griffin I won't write the equations and solve them for you, but I have added a section describing the connection between the spatial and temporal frequencies of the solutions. $\endgroup$
    – mwengler
    Jul 3, 2012 at 7:07

http://www.ftj.agh.edu.pl/pl/000.html?plik=video/jf2008_noc_gitara.html - fresh graduates talks about physics of guitar in polish, some charts could be helpful for english speaking


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