2
$\begingroup$

When a string with fixed ends vibrates (e.g. plucking a guitar string) Fourier Theorem says that the vibration can be expressed as a sum of its normal modes, which are sinusoidal vibrations with frequencies that are all integer multiples of the fundamental frequency.

My question is really simple: since the resulting vibration is a sum of a large number of simple vibrations, with what frequency is the string really vibrating (since, after all, it is a vibration)? Are all the points in the string vibrating with the same frequency but with amplitude modulated in space as happens with the normal modes?

I'm not asking which pitch do we perceive (the fundamental frequency), but at what frequency is the string vibrating.

Improve this question
$\endgroup$
  • $\begingroup$ What exactly do you mean by the frequency the string is "really vibrating"? You know that it contains Fourier components with multiple frequencies, and that the pitch we perceive is determined by their fundamental frequency. So what else do you want to know? $\endgroup$ – knzhou Dec 13 '18 at 23:15
  • $\begingroup$ Yes, what @knzhou said. See: sciencemadness.org/talk/… $\endgroup$ – Gert Dec 13 '18 at 23:30
  • $\begingroup$ For example: the string could vibrate exactly in the fundamental mode. In this case its frequency would be f. If it was vibrating on the pure second harmonic, its frequency would be 2f. Now, for a complex oscillation where many of these modes are superposed, the position, velocity and acceleration of each point will repeat periodically... So it must have "a" frequency. Doesn't it? $\endgroup$ – Pablo Dec 13 '18 at 23:31
1
$\begingroup$

The content of Fourier Theorem is that every periodic function of period T can be represented as a series of (in principle infinite) harmonics, i.e. harmonic motions of frequencies of the kind: $$ \nu_i = i \cdot \nu_1 ~~~~~ i = 1, 2,3 \dots $$ each with its own phase.

Therefore, although made by many different frequencies, the signal which is the synthesis of all its harmonics, as a function of time is a periodic function of frequency $\nu_1$. An example of time variation for a signal made by five harmonics is the following (I have obtained it using the applet in the page http://www.falstad.com/fourier/; it is a good applet and I recommend to play a little with it to get a first understanding how different harmonics do combine; in the abscissa there is time ):

sum of 5 harmonics

The plot contains almost 3 periods of the fundamental mode. From the plot is equally clear that the superposition of normal modes is a periodic function, but definitely it is not a harmonic oscillation.

Improve this answer
$\endgroup$
  • $\begingroup$ Thanks for the explanation; it helps me a lot. A couple of questions: if we were talking about a string vibrating with the harmonic components of your example, the graph would be the displacement that a fixed point in the string would have. Is that correct? Would all the points on the string vibrate with that same pattern? And, since this vibration is periodic with frequency equal to the fundamental, is it correct to say that the string is vibrating at frequency=the fundamental but that it is not a harmonic vibration? Thanks! $\endgroup$ – Pablo Dec 14 '18 at 10:33
  • $\begingroup$ Question #1: no, in general it is not true that all the points will show the same time behavior. Each harmonic above the fundamental has some spatial nodes. If you sit there, there is no contribution to the displacement from that mode and so, in general also the corresponding frequency will be missing. Question #2: in some sense is matter of definitions, but from the mathematical point of view is acceptable. Probably, a more precise description remains that of a superposition of harmonic vibrations with different frequencies. $\endgroup$ – GiorgioP Dec 14 '18 at 14:29
3
$\begingroup$

We perceive the fundamental, the lowest frequency. Even though there are other harmonics present in the vibrating string when no damping is present the shape will move across the string and come back to its initial configuration at the fundamental frequency. The other harmonics are perceived as tone. For example plucking a string closer to one of the supports (a.k.a. boundaries) will create more high end harmonics and in guitar speak that sound "twangy" or "hot", "bright". Plucking near the middle accentuates the fundamental and sounds "warm", or "smooth". But, all are perceived as the same note. Experienced musicians are trained to hear some of the harmonics, once you are trained to hear them you can't un-hear them and music is never the same. The octave is pretty easy to hear but each note contains (at the very least) a major triad in just tuning. So every note you play creates a soft major chord. Very strange from a music theory point of view.

Improve this answer
$\endgroup$
  • $\begingroup$ I understand that we perceive the fundamental frequency as pitch and he other harmonics as timbre. However, my question is not about perception but about the physical oscillation of the string. $\endgroup$ – Pablo Dec 14 '18 at 7:39
  • $\begingroup$ According to what you are saying, it doesn't matter to add many different oscillations with different frequencies: the frequency of the complex is going to be the fundamental frequency. Let's use a silly example: suppose a string vibrating with this superposition: mode 1 with amplitude 0001 and mode 2 with Ampitude 0.999. Is this string vibrating with the frequency of the fundamental tone?? Do we perceive frequency 1 as fundamental? $\endgroup$ – Pablo Dec 14 '18 at 7:46
  • $\begingroup$ Your question seems to convolve more than one idea. First, each point will oscillate at all frequencies present. The motion will be a linear superposition of all modes not one. When you say, at what frequency does the string vibrate you are implying "at what frequency does the shape repeat itself". That will be the lowest frequency because all modes need to come together with the same initial conditions. So the shape repeats at the lower frequency. As for the lowest mode being weak? In that case it may not be noticeable but it is there and the comments hold. $\endgroup$ – ggcg Dec 14 '18 at 12:19

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.