Caution: I have studied single-variable calculus, but have never taken a physics course.

Vi Hart demonstrates with a violin that on a violin string's length, playing at:
1/2 produces a note an octave (a.k.a. P8) higher,
and 1/3 a perfect fifth (P5) higher.

  1. The above causes me to conclude the following in red. Where have I erred?

  2. What are the correct steps to the right answers, per the following?

enter image description here

[5:21] Splitting the string into thirds creates a P5 on the longer segment[,] and a P5 plus an octave on the shorter segments since it's only half the length.

[5:33] The 2 sides of the subdivision vibrate at different speeds. The shorter the string the faster the vibration and the higher the pitch. So the longer right hand side vibrates an P8 lower than the left hand side since the 2:1 equals an P8. This works mathematically also. Dividing the string into 3rds means that each of those 3rds will vibrate 3 times faster than the original or 786 Hz. This is what the short side of this subdivision sounds like. The long side of the string is twice as long so it sounds an P8 lower at a frequency 393 Hz, that is 1/2 the value of the short side, or 3/2 the value of the original.

  • 2
    Maybe it goes wrong at the beginning where you say "each of the 3 wavelengths = 1 P5" - I'm not quite sure what you mean by that - can you rephrase that in other words? (You can't in general equate a certain fixed length of string with a certain interval difference; It's a certain ratio of length difference that will produce a given interval.) Feb 11, 2017 at 22:23
  • @topomorto I meant that each wavelength produces 1 P5. But I think that your parenthesis diagnosed my problem?
    – user26407
    Feb 13, 2017 at 1:47
  • @topomorto I updated and corrected the picture.
    – user26407
    Feb 13, 2017 at 1:57

6 Answers 6


This is the fundamental mode of a vibrating string:

Animation of string fundamental mode

This is the 2nd mode, consisting of two sections vibrating at twice the frequency (aka an octave higher):

Animation of string 2nd mode

The reason this has twice the frequency is simply that the length of each section is only ½ of the string, thus the vibration's momentum needs only half the time to get from one end to the other.

This is the 3rd mode, consisting of three sections vibrating each at thrice the frequency (aka an octave an a fifth / aka a tritave aka a twelfth higher), for the same reason:

Animation of string 3rd mode

All of these work because every node seperates sections which behave the same, thus the forces on each side balance at all times.

Now what you describe seems to be a division into two unequal sections, like this:

Speculative mode of a string vibrating in two different modes in two sections

Here the section on the left-hand side has again thrice the fundamental frequency, for the same reason as with the pure third mode. The RHS has twice the size of the sections of the third mode, so its frequency is ³⁄₂ the fundamental – simply a fifth up.

But it's quite visible here why this isn't actually a mode of a freely vibrating string: the node is not force-balanced on both sides but has most of the time a kink. To uphold this mode, you would need to have a solid anchor in the middle, which is actually possible with a bottleneck slide. But without such an obstacle, the kink would immediately give way as soon as both sides start to get out of sync. The vibration would thus actually look like this (shown with a pause to highlight the original waveform)

What the “seperated modes waveform” would actually evolve like

You see here that the “individual modes” in each section actually mix to a mode on the whole string. But that mixed mode has neither a frequency of ³⁄₁ nor ³⁄₂ the fundamental, rather it is a particular superposition of all the integral multiples of the fundamental:

Mode decomposition of the “seperated modes waveform” evolution

Here's the code used to generate the animations. You need the free Haskell platform to run it.

#!/usr/bin/env runghc
{-# LANGUAGE PackageImports #-}
import Data.Colour.Names
import "dynamic-plot" Graphics.Dynamic.Plot.R2

sinn :: Double -> Double
sinn = sin . (*pi)

coffs :: [Double]
coffs = [2*sum [sinn (k*t) * fdivd t | t<-[0,0.01..1]]/100 | k<-[1..10]]

fdivd :: Double -> Double
fdivd x | x < 1/3 = sin (3*pi*x)/2 | otherwise = -sin (3/2*pi*(x-1/3))

main = plotWindow $ [plotDelay (1/20) . plotLatest $ cycle
            [ lineSegPlot [(x, c * sinn (k*x) * cos (pi*k*t)) | x<-[0,0.02..1]]
            | t<-replicate 20 0 ++ [0,1/20..4] ]
          | (k,c) <- zip [1..] coffs]
   ++ [plotLatest [plotDelay (1/20) . tint white . continFnPlot $
          \x -> sum [c * sinn (k*x) * cos (pi*k*t)
                    | (k,c) <- zip [1..] coffs]
                 | t<-cycle $ replicate 20 0 ++ [0,1/20..4]]
  , forceXRange (0,1), forceYRange (-1,1)]
  • 1
    Wow! +1 for the pretty pics at least! They should explain to the OP where the errors are, in pretty straightforward manner. I tried it with pencil and paper. If I live long enough I'll post it...
    – Tim
    Feb 12, 2017 at 18:12
  • +1. Thanks. I am too ignorant in physics: please allow me some time to cogitate your answer.
    – user26407
    Feb 13, 2017 at 1:49
  • Haskell, eh? You're a daring programmer :-) . Then again, I'd have used R , so there you go. Feb 13, 2017 at 12:54

The problem is at the beginning, when you say:

"each 1/3 of the length of the original string = 1 P5".

This is wrong, because you can't equate a fixed, absolute length of string with a certain interval. an interval can be related to a percentage or ratio difference in length between two string lengths, not an absolute difference in length.

So any diagram where you have a length marked as an interval will lead you to wrong results.


There's another point that the other answers haven't specifically mentioned. A "standing wave" is always the superposition of two (or more) travelling waves. Energy is travelling from one end of the string to the other and back. You only get a standing wave when the travelling wave is an integral or half-integral fraction of the string length (depends on whether the ends are 'closed,' i.e. hard stop, or 'open', i.e. flapping in the breeze). When the two travelling waves have nodes and antinodes at the same locations, you get the standing wave pattern. Remember, the string at an antinode isn't always sitting ata maximum displacent: it's vibrating with maximum amplitude there.

So when you create a node by touching the string at some point, you are requiring the travelling waves to have a node where you are touching in addition to nodes or antinodes at the end of the string. That's why you can't get the asymmetric "1/3 on one side, 2/3 on the other" pattern -- it isn't self-reinforcing and will decay immediately.

  • commons.wikimedia.org/wiki/… potentially useful graphic
    – Dave
    Feb 13, 2017 at 14:07
  • @Dave that is a neat bit of video. But because it's showing a driven system, it might be a bit confusing. Feb 13, 2017 at 14:44

The diagram is completely wrong. In fact, the relationship that you're trying to draw between the division of the string and the notes is fundamentally flawed.

A string plays a single fundamental frequency at a time. How it's divided up determines the frequency played. So a sounding length might play a certain interval above the pitch played by the open string, but that does not mean that the string when divided up is somehow divided into intervals that add up to some kind of interval.

There's also a difference between creating a node 2/3 of the way along a string and then sounding the whole string versus shortening the sounding length to 2/3 of its length. 2/3 of the total length is 3/2 of the open note frequency which is a 5th (almost) above the note played by the open string. Making a node 2/3 of the way along the string and sounding the whole string divides the string into three sections that all together sound at 3 times the open string frequency which is almost an octave and a fifth above the open string.

The section of the string marked "Long" is not sounding at 3/2 the open string frequency. It is sounding at 3 times the open string frequency! There would have to be no node on the "long" section for it to be sounding at 3/2 the frequency.

Also, you can't add sounding lengths and correlate that to adding intervals. The "long" section in no way at all represents either an octave or 2 fifths. It doesn't work like that at all.

  • 2
    This sounds a bit aggressive. The OP did say that they didn't study any physics which would imply that he doesn't know what a stationary wave, a node or antinode is. Feb 11, 2017 at 23:33
  • @xerotolerant I didn't use the terms "stationary wave" or "antinode", and your answer defines "node" and the asker could comment asking for clarification on what I mean by "node" or look up the word "node". So, I feel comfortable using that word without clarification. If the asker has trouble with it, I think they have the rep to comment. Feb 11, 2017 at 23:38
  • From watching the linked presentation, I don't think the crossing points on the diagram were intended to represent nodes at all - they're just a way of visually dividing the string into thirds. Feb 11, 2017 at 23:45
  • @topomorto My intention was to discuss what happens when a real violin string is really sounded in two different ways: One way is to shorten the string by pressing it to the fingerboard. The other way is use a finger lightly applied to the string to create a node which prevents lower frequency vibration modes. That is the meaning of node I was thinking of. Beyond that, the only reasonable way to interpret the diagram is as that of a standing wave, and those points where the wave crosses the x axis are called nodes. So the intention is irrelevant, nodes were drawn. Feb 11, 2017 at 23:49
  • 1
    "not a diagram of a string vibrating at 3/2 the frequency of the open string" - I don't think the pictured shapes are supposed to represent that - I think it's just a way of visually "Dividing the string into 3rds", as per the narration. I agree it looks confusing. Feb 12, 2017 at 0:08

I think the issue here is your understanding of physics and stationary waves. What you have when a harmonic is plays is indeed a "stationary wave".

Where the finger is placed causes "node" to form which in this type of wave is a location of no movement. This would be where the 1/3 mark is. An antinode is the location of the greatest movement where the crests are.

The wavelength of that wave is equal to the side you labeled "Long".

The short side does not have a wavelength of its own. It is part of the whole wave. The short side is not any interval on its own.

This site explains it in greater detail.

Let's look at it using actual numbers

Take A4 = 400Hz

and v = fλ


  • v = speed of sound
  • f = frequency of the sound
  • λ = wavelength

We are idealizing so we are ignoring the effect of tension on frequency of a string by considering the speed of sound in the air. using a little algebra we see

f = v / λ since the speed of sound in air is constant we then see that

f is inversely proportional to λ, meaning when the frequency is multiplied by a number the wavelength is divided by the same.


If we manage to produce an A4 on a string 1 unit long. In order to double the frequency (or produce a P8 over the open string) we have to divide the length of the string by two.

to get two octaves above we can divide by 4.

Now what happens when we divide the string into thirds?

Well since it is inversely proportional we get a frequency of 3 times higher.

E6 = 1320Hz. This is the pitch when played as a harmonic. In which case the new wavelength is actually 2/3 of the open string.

Remember the wavelength for a stationary wave is not the distance between two nodes but double that.

So your errors are as follows:

You are considering the short section as a separate "wave" with it's own length. it isn't.

The wavelength of the wave is actually the the length of the section that you labled Long.

There is no harmonic that is not at least an octave away from the fundamental.

I understand why one might thing a dividing by 3 would produce a P5. When a string is stopped only one side is allowed to vibrate. In addition, no node is produced. So if someone stops a string and plays it one third the distance from the nut (the point labled 1/3 in the image) then resulting wavelength is actually double that of the harmonic on the third. And remember if we double the length we half the frequency so the resulting pitch is E5.

However if were stop the string at the node nearest to the right of the string in your diagram, we get the pitch that matches the harmonic.

  • Though the diagram looks like common harmonic diagrams, I didn't think the question actually related to harmonics at all? Feb 11, 2017 at 23:26
  • I am not sure myself. Now that I think about it lol. But since the OP had a link to an acoustics course and mentioned math I figured an explanation in terms of harmonics would be in order. In the end I explained how that factors into stopping strings which kinda seemed like what the video he was watching was attempting to teach. Feb 11, 2017 at 23:30
  • @topomorto I updated and corrected the picture.
    – user26407
    Feb 13, 2017 at 1:58
  • I don't think the picture needed to be corrected. Unless you don't think it explained what you were thinking when you asked the question. The answers might end up being random looking to someone who reads them in the future if the question is changed too much. Even though you changed it it still is pretty incorrect. Feb 13, 2017 at 2:07

Having spent a long time studying the question, and answers, the OP may have studied physics/maths, but maybe not music. One of the OP's premises is that two P5s = P8. That isn't so. Two P5s = maj10. Possibly the thought is that a P5 is exactly half an octave - P8. Half an octave is a tritone, which musically often doesn't sit well in harmony. So half an octave, scientifically/mathematically is dim5.

The diagram is confusing (confused?) showing nodes, which do relate to splitting the whole string into equal parts. And whether one presses or touches - (to produce a harmonic)- the central node, the new note will indeed be an octave (P8) above the fundamental. However, only pressing it at its 1/3 point will produce the P5 sound; only touching it at either of the 1/3 nodes produces the P5 sound an octave higher: a 12th interval. Now, I'm confused...

  • I think he understands that 2 P5's don't result in an octave. I suspect that he accepts that that is incorrect but doesn't understand the flaw in his steps. Sort of like those 1 + 1 = 3 proofs that students ask about from time to time. At least I think so, though the OP seems to be rather quiet. Feb 12, 2017 at 18:04
  • @xerotolerant I updated and corrected the picture.
    – user26407
    Feb 13, 2017 at 1:58

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