3
$\begingroup$

to be quite honest, I have no idea where to start on this problem mathematically. However, it struck my curiosity and would love to know how it works on a mathematical level.

the problem

You have a guitar. All of the strings start at a resting position. When you play a string, it is now oscillating between two points (as far as I can tell). The speed at which it oscillates is correlating to the frequency, and the distance between the points is how hard the note was struck or the volume of the note.

You play the E string, open, and let it ring. Then you play the 7th fret of the same string, and let it ring. Now how does one determine how quickly the string is oscillating between the two points in this scenario, in a unit of measurable distance?

All I really have so far is that E and B are a perfect fifth apart, making the ratio of their frequencies 3:2. If one were to play the 12th fret of the same string, that would be a 1:2 relationship. Therefore it follows that the speed they oscillate should follow the same proportions. I, however, am not sure how to get an initial distance.

Edit for some clarity; the application

to put this question in context, I am trying to programmatically work with a three-dimensional model of a guitar as it is played by a guitar player and make it visually convincing.

As such, when he plays an open string, it would move more slowly. If he put his finger on a fret on that string and played it, nothing from his finger to the nut would vibrate, but the rest would vibrate FASTER. I'm trying to discern a reasonable level of accuracy of this visual effect.

Thank you

$\endgroup$
  • $\begingroup$ Meanwhile take a look at this or any other good physics book. I don't understand your description completely, but it seems to have some misunderstandings (e.g. "oscillating between two points" is not correct). $\endgroup$ – Szabolcs Jun 19 '14 at 18:30
  • $\begingroup$ in what sense? Perhaps I used the wrong terminology, but it is vibrating, so moving back and forth $\endgroup$ – Crow Jun 19 '14 at 18:35
  • $\begingroup$ It can vibrate in many different ways. This type of vibration cannot be described as the oscillation of a single variable. Typically, when analysing this problem, it is decomposed into the superpositions of vibrations that look like these. You'll have this all sorted out after the question is migrated. $\endgroup$ – Szabolcs Jun 19 '14 at 18:39
  • $\begingroup$ That'll have to be a visual representation with pretty crazy FPS count, so people could "see" the speed. The eye itself isn't fast enough. What you should be more concerned about is amplitude, then just blur the string over that range, implying that it vibrates faster-than-visible. That'll look far more convincing than an obviously-too-slow simulated vibration. $\endgroup$ – leftaroundabout Jun 20 '14 at 0:13
2
$\begingroup$

Update: I guess I misunderstood the question. What I answered was "how fast do points on two strings move in relation to each other".

Answer: The relative speed does not stay constant, but oscillates with the frequency equal to the difference in the frequencies. Sometimes the points on the string are in phase and the relative speed is zero but after a while they move exactly opposite to each other.

In addition one can not talk just about a single frequency in the case of any musical instrument. There are always side bands. This is because off the material the strings are made of. The different frequency sound waves move with different speeds - this is called dispersion. Due to this you have more than one solution to the wave equation and it is said to have many modes exited. So not even a single string oscillates with a simple relation for the speed. This is also why different instruments have a different sound even when playing the same note. Maybe you have heard a frequency generator make pure tone and the lack of side tones is what makes it sound boring or even annoying.

To give you a feeling of what is going on here are few plots. First lets show two sine waves which have frequency ration of 2/3. I plot them with Mathematica. If you happen to have one you can copy the code and play with it.

f1 = 2/(2 \[Pi]);
f2 = 3/(2 \[Pi]);
p1 = Plot[{
   Sin[t f1],
   Sin[t f2]
   }, {t, 0, 100}]

two sine waves

You can see, that the relation between changes in time. Now lets look a sine of which frequency is the difference and a difference the two sines on the previous plot.

p2 = Plot[{
   Sin[t f1] - Sin[t f2],
   Sin[t (f2 - f1)]
   }, {t, 0, 100}]

sum of freq and dif of sines

You can see, that the constructive interference or so called frequency beating happens with a frequency of the difference of the two of the initial sines. It is also called the frequency mixing. This is what makes some accords on some instruments sound really nice... the combination of frequency mixing, higher harmonics and side tones together make a complicated spectra of frequencies.

If you want to see the speeds instead of the positions over time, then you need to take a derivative by time. This gives you cosines and the things you get would have the same properties.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

"How does one determine how quickly the string is oscillating between the two points in a unit of measurable distance?"

The measure of how quickly the string is oscillating is its frequency, stated in Hertz (cycles per second).

The frequency of played string is can be found by

enter image description here

Ref: Piano key frequencies

where n is the musical key number, e.g. middle C (C4) is 40

Or alternatively

enter image description here

Ref: Note frequency

where n is the number of half-steps away from middle A (A4), e.g. for middle C (C4) n = -9

See also Guitar harmonics.

Raising an octave doubles the frequency, i.e.

C4 = 261.6 Hz
C5 = 523.2 Hz

G4 is a perfect fifth above C4 and is approximately the same number of hertz above C4 as C4 is above C3.

enter image description here

Further elaboration

First octave

c3 = 2^(-21/12.)*440 = 130.8 Hz

Second octave

2 x c3 = 261.6 Hz

c4 = 2^(-9/12.)*440 = 261.6 Hz

Perfect fifth

3 x c3 = 392.4 Hz

g4 = 2^(-2/12.)*440 = 392.0 Hz

Third octave

4 x c3 = 523.3 Hz

c5 = 2^(3/12.)*440 = 523.3 Hz

Major third

5 x c3 = 654.1 Hz

e5 = 2^(7/12.)*440 = 659.3 Hz

As I recall the discrepancies are due to the Pythagorean Comma.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't think that's what he's asking - speed vs frequency depends on amplitude :-) $\endgroup$ – Carl Witthoft Jun 19 '14 at 19:57
  • $\begingroup$ @Carl - He asks for distance (between the end points?) but I think he wants frequency if he's talking about oscillation. Otherwise he might loosely mean distance of amplitude but that would just affect volume, not pitch. $\endgroup$ – Chris Degnen Jun 19 '14 at 20:21
  • $\begingroup$ my question is a bit less abstract than people are making it out to be -- I mean as in physical, measurable distance which the string travels when it is hit, less so than the effect of sound. Eg, when you play an open E, it should go back and forth slower, but if you played the 7th fret on the same string, it should be moving faster $\endgroup$ – Crow Jun 19 '14 at 22:29
  • $\begingroup$ The distance the string moves, i.e. its amplitude, will depend how hard you pluck the string and that will only affect the volume. The pitch is controlled by the string length. The actual wave shape of a plucked string is shown in a video here: Motion of Plucked String $\endgroup$ – Chris Degnen Jun 20 '14 at 0:52
  • $\begingroup$ Another interesting video here: Super Slow Motion Guitar Strings $\endgroup$ – Chris Degnen Jun 20 '14 at 1:45
0
$\begingroup$

how fast is a stringed instrument's string interpolating?

I assume you mean vibrating. Notes that are 1 octave apart vibrate at double the frequency (i.e. the note A should have the frequencies of 55, 110, 220, 440, 880, 1760, 3520, 7040 and 14080 Hz). The octave is divided by 12 to get our semitone scale. The frequency difference for each step is based on an exponential formula.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Well, that's its frequency, but not the speed of a differential length of string at the location of peak amplitude (but of course at zero displacement, where the oscillation is at its fastest speed). $\endgroup$ – Carl Witthoft Jun 19 '14 at 19:56
0
$\begingroup$

The other answers have related translation of a note into a frequency.

For a guitar, I think that you wish to know the position of each point in the string as a function of time.

Let the length of the string be L.

Defined the point at which is plucked P along the length, with amplitude A.

The initial position is a triangle (not necessarily isosceles)

   y(x,0) = xA/P, x <=P or (x-P)A/(L-P), x >= P

This then needs decomposed into individual frequencies using Fourier decomposition. Each of these Fourier components then oscillates at a different frequency to obtain the full time dependent profile.

Well, I was about to elucidate all that, but stopped myself. See for example Power point from BYU Physics Dept., MathWorld

enter image description here

In the above, v is the speed of the wave, which in turn determines the frequency via the standard wave equation.

All this I recall from lectures years back. The more tricky bit is: how to the amplitudes decay? The following paper has some discussion on this:

www.lps.ens.fr/~laetitia/WAVESTRING.PDF

| cite | improve this answer | |
$\endgroup$

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.