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When you energise a taut string, the following resonant modes of vibration occur:

enter image description here

Plotting on the frequency domain, you can see their corresponding frequencies:

enter image description here

But what is the underlying physical principle? Why does this happen?

Is there any way of explaining it that could be understood by a smart 15 year old?

EDIT: I'm going to give my best attempt so far. Here goes:

  • We can start with sympathetic resonance. Sounding a particular frequency, a pure sinewave. And noticing the string resonates sympathetically at the frequency of each harmonic. Say this is explained and understood.

  • Now imagine that plucking a string is equivalent to a burst of white noise, which contains frequencies all the way through the spectrum. This could be approached backwards, by starting with random frequencies and noticing that resultant wave produced looks like white noise.

If the above is scientifically correct, then it restricts the domain of the question.

I would really like to be able to understand it scientifically and also be able to explain it intuitively.

PS Images from http://www.embedded.com/design/real-world-applications/4428811/2/Building-an-electronic-guitar-digital-sound-synthesizer-using-a-programmable-SoC

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    $\begingroup$ I'm not sure if I understand the question. These are simply solutions of the wave equation that fulfil the boundary conditions $\endgroup$ – pfnuesel May 8 '14 at 9:46
  • $\begingroup$ For a not-so-mathematical explanation, see this other question: music.stackexchange.com/questions/5489/why-do-harmonics-happen $\endgroup$ – Antonio Ragagnin May 8 '14 at 10:00
  • $\begingroup$ Do you understand the mathematical derivation? $\endgroup$ – jinawee May 8 '14 at 19:39
  • $\begingroup$ @jinawee, I've looked through the video linked by JamalS, so say I have some partial confidence in the wave equation. But it still seems a world away from understanding the phenomenon intuitively. $\endgroup$ – P-i- May 8 '14 at 21:50
  • $\begingroup$ If I understand your question correctly, you are not asking why these resonant frequencies exist, but rather why does a pluck excite all of the resonant frequencies. Is that correct? If so, see my answer below. $\endgroup$ – Chris Mueller May 8 '14 at 22:27
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When you release the plucked string, its shape is momentarily triangular: tied down at the ends and pointed at the location of your finger. But the solutions to the wave equation are not triangle functions, but sinusoidal functions, whose displacements from rest obey $$y_n(x) \propto \sin \frac{2\pi x}{\lambda_0 / n},$$ where $\lambda_0$ is twice the length of the string. These waves, whose frequencies are integer multiples of the fundamental frequency, are the harmonics.

There is a theorem that you can add up all of these well-behaved $y_n(x)$ and generate any shape $y(x)$ for the real string that you want. The subject is called Fourier analysis. And that's just what happens when you release your guitar string. From the string's perspective you've just excited a whole bunch of different modes with different $n$, and they all begin to oscillate at their own frequencies.

It's worth pointing out that you have some control over which harmonics you excite by choosing where you pluck the string. Here's how the harmonics up to $n=16$ (four octaves above the fundamental) contribute to the shape of a guitar string plucked near the middle, near the sound hole, and near the nut:

guitar harmonics

The "exact" triangle shape is in blue; the fundamental excitation is in green; the fundamental plus first harmonic in red, then cyan, magneta, yellow, etc. as more harmonics are included. Plucking a guitar string near the nut (bottom figure) excites lots and lots of the higher harmonics. This is a thing you can hear on a guitar: strumming close to the nut produces a harsh, pinched sort of an "eeee" sound. By contrast, if you pluck the guitar string very near the center of the string, you put very little energy into the 1st, 3rd, 5th, harmonics, which have a node at the middle of the string. This gives a sort of rounder, "oooo" sound to the strings. Give it a try!

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    $\begingroup$ Triangle functions ARE a solution. But sinusoids are the separable solutions (they would be the eigenfunctions of the Sturm-Liouville problem). $\endgroup$ – jinawee May 9 '14 at 8:55
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When you pluck a string it does not start out like the fundamental above. The string is pulled into a bent shape of two straight lines and an angle and it may not be bent at the middle.

Releasing the bent string causes a bunch of harmonics of various amplitudes depending on how far off-center it was bent. (It can not return to the bent angle shape and the energy has to go somewhere). The result from that shape is all harmonics and sounds like a "rich" sine wave from the odd harmonics dominating.

A guitar or violin is plucked very off-center so it is more like a sawtooth and gets all harmonics, odd and even with a set of amplitudes distinctive to the instrument.

This was first(?) studied in any detail by a French monk named Mersenne who used long heavy wires between fence posts in order to get vibrations slow enough to count.

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The response can be derived mathematically. Let $u(x,t)$ denote the displacement of a point along the string at $x$ at time $t$. The function obeys the wave equation in flat $d=2$ Minkowski space,

$$\frac{\partial^2 u(x,t)}{\partial t^2} - v^2 \frac{\partial^2 u(x,t)}{\partial x^2}=0$$

If we pinch the string at the middle, this corresponds to a condition on the configuration of the string at the initial time, i.e. it determines $u(x,0)$:

$$u(x,0)= \begin{cases} 1-x, & x\in[0,1]\\ 1+x & x\in[-1,0] \end{cases}$$

Furthermore, we must impose Dirichlet boundary conditions as the string is fixed at either end, i.e.

$$u(0,t)=u(l,t)=0$$

to ensure movement at $x=0,l$ is prohibited. Solving the wave equation via Fourier series is tedious but easily doable. Eventually, we obtain the wave harmonics, images of which are available in the OP.


A simple graph of the initial condition $u(x,0)$:

enter image description here

We could specify any $u(x,0)=f(x)$, or solve for a general configuration.

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  • $\begingroup$ Sorry I cannot yet upvote. One problem I have here is that it is explaining a tangible phenomenon in terms of equations that are not obvious. So the question becomes "why do these equations hold?" Is going through the equations going to be the best path for an intuitive general understanding? $\endgroup$ – P-i- May 8 '14 at 15:25
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    $\begingroup$ @P-i-: The mathematical approach (in my opinion) is the most rigorous to understanding the phenomenon, but physical arguments may help elucidate it. See the video: youtube.com/watch?v=r2GIY2ZmXPY for an excellent undergrad derivation of the wave equation, which should fully justify why it holds to you. $\endgroup$ – JamalS May 8 '14 at 15:30
  • $\begingroup$ Isn't the amplitude in your example too large? Shouldn't you have to consider nonlinear effects? $\endgroup$ – jinawee May 8 '14 at 19:38
  • $\begingroup$ @jinawee: Can you elaborate? $\endgroup$ – JamalS May 8 '14 at 20:46
  • $\begingroup$ @jinawee of course it's an exaggerated picture. Physically the y axis is just in another units. Mathematically, it makes no difference for wave equation since it's linear. $\endgroup$ – Ruslan May 9 '14 at 5:44
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When you pluck a guitar string the potential you apply to the string is approximately a Dirac delta function. That is to say, the release of the string is a near instantaneous kick. One of the beautiful properties of the delta function is that its Fourier transform is unity. This means that it is made up of equal components of all frequencies. So, when you pluck the string you excite every single resonant mode equally (in the delta function limit).

What determines the different sounds of different instruments is how long each resonant frequency can be sustained, i.e. the $Q$ of each resonant mode. Your second plot shows these different $Q$ values nicely. The $Q$ is roughly proportional to the width of the peak at each frequency where a higher $Q$ value means a narrower, taller peak. Resonant modes with a wider spread (lower $Q$) will die off more quickly as they transfer energy to the support structure and heat in the string.

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A couple things play in here. First, the string is "closed" at both ends, meaning the ends are locked down and can't move. This means any resonant wavelength must have "nodes" , which is a contraction of "no displacement," at the ends. By comparison, the acoustic waves in an open-ended tube may have a node at one end, but the other end is unrestrained and could be a maximum. Next, the resonant "standing wave" for each frequency is actually the combination of travelling waves moving in phase and in opposited directions along the string. The fundamental harmonic has a maximum half-way down the string; the next harmonic has two maxima at 1/4 and 3/4 length plus a node at 1/2 length. And so on for all higher harmonics. Note that you can suppress, e.g., the fundamental by placing your finger at the 1/2-length point to force a node there. This will in fact suppress all the odd-numbered harmonics which have a maximum at 1/2 while allowing the even-numbered harmonics to continue to propogate.

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    $\begingroup$ That (folk?) etymology for 'node' looks dubious to me. These two etymologies suggest a different origin, though it's hard to tell when and how the term came to acquire that specific meaning. $\endgroup$ – Emilio Pisanty May 8 '14 at 11:30
  • $\begingroup$ Will a string plucked in the center have even harmonics? $\endgroup$ – C. Towne Springer May 8 '14 at 11:57
  • $\begingroup$ @C.TowneSpringer No matter where you pluck a string, only the resonant frequencies will survive. Think of the "pluck" as an impulse waveform which contains a whole ton of frequencies. It is true that plucking at the maximum amplitude location of the fundamental will produce a cleaner sound (due to less "breakup" of the pulse). $\endgroup$ – Carl Witthoft May 8 '14 at 20:00
  • $\begingroup$ @CarlWitthoft I was thinking plucking at the center point will give only odd harmonics. Maybe I'm confusing it with a time dependent function of the same shape. $\endgroup$ – C. Towne Springer May 8 '14 at 20:43
  • $\begingroup$ @CarlWitthoft "only the resonant frequencies will survive": where would others go if they are initially there? And how can they be initially there given the boundary conditions? $\endgroup$ – Ruslan May 9 '14 at 9:39

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