We learn that waves travels in strings under tension, have fundamental frequencies, but I have no luck understanding why don't musical instruments have simple strings with uniform thickness which we learn about in waves. \begin{gather*} \nu = \frac{nv}{2L} \end{gather*} where $\nu$ is the frequency of vibrations on a string, $n$ is some natural number, $L$ is the string length, and $v$ is the wave velocity.

So, why do strings musical instruments have spring type structure, what would be different if they were simply wires of uniform thickness?

  • 14
    $\begingroup$ That's just the helical coating, a practical way to give a string a certain mass per unit length and protect the string core that is responsible for tension. Lighter steel strings or nylon strings do not have this helical coating. For dofferent kinds of winding types see "String construction" here: en.wikipedia.org/wiki/String_(music) $\endgroup$
    – Quillo
    Commented Nov 1, 2023 at 10:25
  • 1
    $\begingroup$ @Quillo, The question is, if all you want is more mass, why go to the trouble* of winding the one wire around the other? Why not simply use an unadorned length of thicker wire? (see fraxinus's answer for,... um,... the answer.) [*extra production time+extra tooling=increased cost] $\endgroup$ Commented Nov 2, 2023 at 17:14

3 Answers 3


The important part here is that in a musical instrument, you care about more than the fundamental frequency.

E.g. you want the frequency to be independent from the sound volume in a wide dynamic range and stable over time and use, you want a certain harmonic content of the sound and you would like overtones that are exact integer multiples of the base frequency.

The formula you cite is a good approximation if your string is tensioned well into the linear elasticity region, the string has negligible stiffness and the additional tension from the vibration is negligible compared to the tuning tension.

Stiffness shifts the overtones from their sweet spots, making the sound dull; not enough tension and the pitch starts to depend on the sound volume, making the whole thing untunable. In fretted instruments, too little tension also makes frets out of tune from each other. Too much tension and you bend the instrument's frame.

With a given length of the string, higher pitched strings are easy - you get a certain size of steel or polymer wire, get it tensioned to tune, done.

If you want lower-pitched strings, you have to make them heavier without making them much stiffer - so a thicker string won't do from some point on. But you have to add mass to the string.

Winding a second wire to the first is the go-to method. The resulting string is generally as stiff as the bare wire, but it is as heavy as needed, allowing for the lower pitches.

In regard to the sound waves, the helical shape is of no importance as long as the resulting irregularities are much, much shorter than the highest interesting overtone wavelength.

The wound strings do, in fact, cause problems.

First, the wound wire may disengage from the base wire, developing a quite unpleasant buzzing sound when both wires vibrate somewhat independently and beat into each other. Time to replace the string.

Second, the helical shape allows for dirt to enter into the gaps, changing the mass distribution over the string and shifting the overtones from their pleasant frequencies.

Third, moving fingers along a wound string makes an unwanted squeaking sound.

  • 4
    $\begingroup$ As someone who moonlights playing guitar, this is a fascinating answer! I actually now realize a lot of the subtleties that arise from playing strings over a long period are just consequences of the material (i.e. the squeak from sliding on low strings or the dullness of old and dirty strings). $\endgroup$ Commented Nov 1, 2023 at 9:26
  • 5
    $\begingroup$ I believe the bending stiffness of a rod goes up as the 4th power of the diameter. $\endgroup$ Commented Nov 1, 2023 at 10:52
  • 2
    $\begingroup$ It's also worth noting that "wrap a wire around the string" produces the more common "round-wound" strings, but you can also wrap a something with a rectangular cross section, basically a ribbon, (or at least having a flat outer profile) to get "flat-wound" strings that are smooth that, at least in my experience, lead to much less noise from finger movements and last much longer due to not collecting as much dirt and oil. $\endgroup$ Commented Nov 2, 2023 at 12:16
  • 1
    $\begingroup$ String squeak is not always undesirable and is sometimes perceived to add authenticity or a percussive effect in certain genres. reddit.com/r/explainlikeimfive/comments/2y1nuk/… $\endgroup$ Commented Nov 2, 2023 at 23:56
  • 1
    $\begingroup$ I want to make one correction to this answer: "Stiffness shifts the overtones from their sweet spots, making the sound dull" I assume that "shifting the overtones" is referring to inharmonicity, which does not make the string sound dull but rather a bit out of tune. The "dullness" comes from the damping effect of a thick string on high overtones. $\endgroup$
    – Edward
    Commented Nov 3, 2023 at 0:02

It is much simpler reason for that: strings are wired like that because the resulting texture is easier and more reliable for the music player to grip on while playing. That's all. Smooth strings would randomly slide off from player's fingers and provide bad playing experience. Similar reason to why tires of regular cars aren't smooth, but have a certain texture.

Involving some fancy formulas that involve unnecessarily "smart" terms like the frequency of vibrations, some natural number, and wave velocity, is way overthinking and overcomplicating this.

  • $\begingroup$ Have you ever played a guitar? The thinner strings are unwound. Usually the smallest 3, but occasionally only the smallest 2. $\endgroup$
    – Edward
    Commented Nov 2, 2023 at 23:58

It’s the same reason why transmission wires are twisted: The fundamental transversal proper modes of constant frequency of a simple string come with a continuous spectrum of rotational modes: for each transversal mode of form

$$A (t,x, y ,z) = \sin(\omega_n t) \left( a_x \sin (\frac{n \pi x}{L}) + a_y \dots\right)$$

with a fixed direction of the amplitude, there exists an infinite ladder of spin modes, where the three spatial coefficients perform a rotation in time.

Classically, the rotational modes atop a fundamental are not quantized by (as in quantum mechanics by the integer multiples of $\hbar$), but have a continuous spectrum.

Transferring angular momentum to the simple string by violinists technique, you can make the string whine as you like, well known from oriental music.

In order to suppress the whining of the transversal modes by the stimulation of the rotational degrees of freedom, the simplest way is to add a strong, inner dissipation for rotation in a twisted string on the one hand and to suppress the trivial uniform rotation modes by a symmetry.

Electromagnetically, twisting a pair of wires suppresses the dipole modes of a parallel twin conductor, such that the lowest mode are quadrupole modes. This reduces the loss by radiation to zero nearly and reduces the transfer to an absorbing external medium drastically by a faster decay of the radiation field potential from dipole $1/r^2$ to quadrupole $1/r^4$ in terms of $e^{-\frac{r}{\lambda}}$, typically.

Acustically at 300 Hz and wavelength 1 m, the rotation mode amplitude falls off to zero within some meters for a quadrupole source, for example, a string without a resonance body or an open speaker without any box exhibiting its backwards wave from propagating into space.

  • 3
    $\begingroup$ What kind of rotational modes do you mean? The rotation of the thick string's material around string's length? If so, why is their spectrum continuous if the string is fixed at two ends? $\endgroup$
    – Ruslan
    Commented Nov 1, 2023 at 7:55
  • $\begingroup$ Angle an rotation frequency of a skipping rope is a canonical pair of variables with no geometric quantization, same as top with fixed axis . Rotational modes of a cylinder with fixed ends have discrete spectrum, but that is probably spaced at higher frequncies and couples badly, I guess, from the difficulty to hear anything if I start a twisted string between two fingers acucurately in the straight line. $\endgroup$
    – Roland F
    Commented Nov 1, 2023 at 8:22
  • 1
    $\begingroup$ Re "exhibiting": Do you mean "inhibiting"? $\endgroup$ Commented Nov 1, 2023 at 18:16
  • 1
    $\begingroup$ Transmission wires inherently don't need to be twisted (that is for external reasons. $\endgroup$ Commented Nov 1, 2023 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.