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I have a question about standing waves on strings. I'll try to explain the best I can, I searched and researched the whole day yesterday but I am confused still:

  1. Every frequency has a single, and only one possible wavelength. This is what I come across over and over again, with no other conditions mentioned. eg. E2 is 82.407hz with a wavelength of 4.129m. The wavelength seems to be referenced as a constant, nonchanging measurement. Is this correct?
  2. The above is proven false with simple tuning of the guitar, where the standing wave is clearly the same length (the half-length of it from nut to bridge, from node to node) yet the frequency is changed by increasing tension. The same length of the string produces for example fundamental pitch of the note E2 and F2. Does this prove the above statement false?
  3. Does this mean that the wavelength of the standing wave and the wavelength of the sound that it produces can be different?

EDIT: For simplicity's sake, unless it's relevant, I am talking about the fundamental only. Also, I am assuming that the speed of sound is the same, through the air, same temperature same everything, the string are the same etc.

It is really puzzling to me how seemingly same wavelengths can create different frequency on the same guitar, with the same string. And further how a guitar with a shorter nut-to-neck distance can create the same frequency. Thanks a lot in advance for any help figuring this out.

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6 Answers 6

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You overlooked the fact the speed varies when tension varies. Note that since $v=f\lambda$ a variation in both $v$ and frequency $f$ can mean wavelength $\lambda$ stays constant.

Note that the speed of the wave on a string is given by $$v=\sqrt{\frac{T}{\mu}}$$ where $T$ is the tension in the string and $\mu$ is the mass per unit length (linear mass density).

When you increase the tension in the string, the wave speed increases. This also means that if you decrease the tension, the speed decreases. In either case, the wavelength remains constant.

More specifically, as the tension increases, so does the speed and frequency (so you hear a higher pitch), but the wavelength remains constant. Of course the converse is true, and as the tension decreases, so does the speed and frequency (so you hear a lower pitch), but again the wavelength remains constant.

The point is, even though the string wavelength always stays the same, the frequency need not.

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    $\begingroup$ Thanks Joseph. For some reason I assumed that the speed refers to the speed of sound. I just now managed to wrap my head around the idea that the wave can move 2xfaster through the 2xlonger string than in the short one, yet the peaks of their fundamental will reach the +1 polarity at the same time, all the time if they are of the same fq. If that makes sense. :D Thanks for the explanation! :) $\endgroup$
    – Dimitri
    Aug 4 at 7:48
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    $\begingroup$ No problems. Good luck with it all. $\endgroup$
    – joseph h
    Aug 4 at 7:49
  • $\begingroup$ Would it be accurate to say that, when applying the formula $v = f \lambda$ to a guitar string, the relevant speed is the speed of sound in the guitar string itself? $\endgroup$ Aug 4 at 22:00
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    $\begingroup$ @TannerSwett doesn't sound correct. The speed of wave, if we take the frequency and wavelength from the OP, is about 340 m/s. And speed of sound in steel is an order of magnitude higher. $\endgroup$
    – Ruslan
    Aug 5 at 18:37
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Your logic and research are sound (pun intended), the last piece of the puzzle is the fact that the speed of sound is different in different materials and it changes when you change pressure/density of the material. To your specific questions:

  1. That fixed reference wavelength you mention is for air at room temperature and sea level density. It will be different when pressure or temperature of the air changes, or if the sound wave is moving through a different material (i.e. guitar string, wall of a house). The equation or relationship between frequency and wavelength only holds when pressure and temperature do not change.

  2. Changing the string tension changes the pressure and density of the string material, hence changing the wavelength-frequency relationship

  3. The wavelength is different in different materials. In your example the standing wave refers to the string, the sound produced is the air, these are two materials with very different densities and hence different wavelengths for the same frequency sound.

Thank you for the well thought out question!

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    $\begingroup$ Thanks for the detailed explanation. :) I somehow got into my head that the speed of sound must be the speed of the wave through the string. $\endgroup$
    – Dimitri
    Aug 4 at 7:54
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    $\begingroup$ @Dimitri The speed of sound isn’t even a constant. It’s a material property that varies with the density of the material the sound is moving through. $\endgroup$ Aug 4 at 17:26
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We have to carefully analyze what a wave is and how it is different from oscillation.

If there is a medium that can store energy in two different ways, like potential energy and kinetic energy, and if these two energies are linked the way that, the more one type of energy contributes, the other changes more rapidly, then you will have an oscillation, that is, energy will be transfered from one form to the other at a certain frequency. An oscillation is seen as this change happens "in a place", even if this place has a certain size. This is true for the string. In a string one energy (tension) is directed along the string, the other one (kinetic energy) is perpenticular to the string. So every single point of the string oscillates at the same frequency with different amplitude. This oscillation can originate from elongation of every single point or from picking the string, in this case a spectrum of harmonics is created, but as higher frequencies are damped in the end there will be just the fundamental in the form of a "standing wave".

So a standing wave is not a wave but the physical implementation of an oscillation.

Now the moving string interacts with the air and then there will be a free wave excited. The frequency of this wave is determined by the string, but the wavelength and so the speed is determined by the properties of the air, in this case bulk modulus and specific mass.

To conclude: a wave and an oscillation differ in the way that the center of energy travels or is fixed. If a pulse is reflected between two mirrors inside the mirror it is seen as a wave, from the outside as an oscillator.

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  • $\begingroup$ Hi Erna, thanks for the answer. Some things I don't understand: ""So every single point of the string oscillates at the same frequency with different amplitude."" How is this so, then the string couldn't generate overtones but only a singular frequency, or am I mistaken? $\endgroup$
    – Dimitri
    Aug 8 at 7:26
  • $\begingroup$ OK Dimitri, you are right. The point is: every model comes only close to reality. As real objects exist in space and time we simplify it by neclecting space when we see oscillations. So if you pick the string you create an excitation that looks like a gaussian peak and can be seen as a set of frequencies with a gaussian amplitude distribution. This excitation then runs along the string, is reflected and finally all frequencies die that do not create standing waves. So the bounded string is the resonator. Prince picked in unseen ways ;-) and excited people. $\endgroup$
    – Erna
    Aug 8 at 9:16
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The source of confusion is that you are comparing the frequency of one wave, to the wavelength of a different wave. That is why they are not inversely related as you expected them to be.

One of the waves plots the maximum amplitude of the string on the y axis, with time on the x axis. For this wave, when you tighten the string, the frequency increases (so the sine wave gets "denser") and the wavelength also decreases, however the corresponding "wavelength" is measured in units of time, and we call it a "period", denoted $T$ in physics. For this wave, we have

$$f = \frac{1}{T}$$

The other wave going on here is just a plot of two physical coordinates. The x axis is the distance along the fretboard, and the y axis is the maximum amplitude of the string. When you tighten the string, the fundamental wavelength of this wave will not change, nor will its "frequency" - here, the "frequency" is in spacial units, and it is more commonly called the "wave number", denoted $k$ in physics. It is the number of wave cycles per unit distance. For this wave, we have

$$k = 1 / \lambda$$

Sometimes people define the wave number in terms of an angular frequency instead, and with $2\pi$ radians in one rotation that would give $k = \frac{2 \pi}{\lambda}$. This can be done with the frequency in time $f$ just as well.

For either of these waves, the wavelength is inversely proportional to the frequency. But comparing wavelengths and frequencies across different waves won't guarantee that kind of relation.

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The feature that is preserved when an acoustic wave transitions from one medium to another is frequency. The frequency of oscillation is the same in the guitar string, the guitar top (and back) and the sound waves radiated into the air.

Mechanically, what's going on is the string is vibrating. This causes minute wobbles in top of the guitar. At the point of contact, the bridge, the motion is in phase -- when the string is pulling inwards, the bridge is nudged slightly that way, when it relaxes the bridge moves away. So they're synchronized in time.

However, the speed of sound in these two media are different. Since $c = f \lambda$ is true for all simple waves, different speeds of sound for the same frequency means different wavelenghts.

The same thing happens with the interaction of the guitar top and the air -- the top is oscillating in some pattern, that impinges on the air, creating pressure (sound) waves that radiate outwards. It this timing, and thus the frequency, that is preserved in this interaction.

So it's not "the same wavelength produces different frequencies" it is the "the same frequency has different wavelengths depending on the material the sound is propagating through".

As an aside, it's useful sometimes to think about guitar strings in the time-domain. By this I mean that the period (inverse of the fundamental frequency) in a string with fixed ends is the time it takes for a small pulse to make a full-circuit (down and back) along the string. To me, it is semi-intuitive that if you increase the tension in the string, then a pulse like that will move faster. At a couple of points in this video https://youtu.be/tz8Rsz-FOHo I can see impact of the striker move up and down the wire; when the guitar string is hit with the pick similar things happen (at least initially)

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This might surprise you, but the wavelength of the standing wave in a string is pretty much always different from the wavelength of the sound it emits. What is equivalent between the two vibrations is the frequency, or, in musical terms, the pitch. How the frequency relates to the wavelength is dependent (in a usually not straightforward way) on the material in which the vibration is propagating. The material of a string is very different, of course, from the material of the air in which the sound propagates. So, even though the two vibrations have the same frequency, they will not at all have the same wavelength.

Regarding your point 1, a frequency does not have one and only one wavelength. Again, the relationship between frequency and wavelength is dependent on the material in which the vibration is occurring. For a string, the most important elements are the string length, the mass per unit length, and the tension on the string. As you indicate in your point 2, changing the tension will change the frequency, even though the wavelength will not change. Given a string vibrating at a particular frequency, it would be possible -- maybe not useful, but possible -- to find another string with, for instance a different mass density, and put a different tension on it so that it vibrates at exactly the frequency of the first string.

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