I've notice that thicker strings have lower frequencies, but is there a specific relationship between theses two?


2 Answers 2


It's a bit more complicated than "thicker" or "thinner," but if you specify that the strings your are comparing have the same length ($L$), have the same volume density ($\rho$), and have the same tension ($T$), then a thicker string will have a lower fundamental frequency than a thinner one.

Now let's look at this with some mathematical tools:

The fundamental frequency of an ideal string (the real stiffness of a string can affect the frequency slightly) fixed at both ends is $$f_1=\frac{1}{2L}\sqrt{\frac{T}{\rho A}}$$ where $A$ is the cross-sectional area of the string of radius $R$: $$A=\pi R^2.$$ If we put this area in the fundamental frequency relation we get $$f_1=\frac{1}{2LR}\sqrt{\frac{T}{\rho \pi}}.$$

This shows us that for strings of same length of the same density experiencing the same tension, the thicker string will have the lower fundamental (and consequently a lower pitch).


A vibrating string is governed by the wave equation:

$$\frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}$$

Where $y$ is a function $y(x,t)$, $T$ is the string tension and $\rho$ the linear density ($\mathrm{kg/m}$) of the string, and:


Solved, this equation yields:

$$y(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$

For $n=1,2,3,4...$ and with $L$ the string's length.

(the $A_n$ are coefficients yet to be determined from the initial condition (state) of the string)

(My full derivation of the solution can be found here)

Now let $\cos\Big(\frac{n\pi ct}{L}\Big)=\cos\omega t$

with $\omega=2\pi f=\frac{n\pi c}{L}$

For $n=1$, i.e. 'the fundamental' frequency:

$$\boxed{f_1=\frac{1}{2\pi L}\sqrt{\frac{T}{\rho}}}$$

It now becomes apparent that the frequency (pitch) of the fundamental is:

1) inversely proportional to the string's length.

2) proportional to the square root of the tension.

3) inversely proportional to the square root of the string's density.

$$All other things being equal, heavier strings will produce lower pitched notes.**

This is somewhat analogous to the simpler mass-spring system, where higher $m$ also leads to smaller $f$.

$$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$$

  • $\begingroup$ I believe you have an extra $pi$ in your $f_1$ relation. Also, $\rho$ has a standard use as a volume density, so seeing it in this relation could be confusing. $\endgroup$
    – Bill N
    Jan 17, 2019 at 16:45
  • $\begingroup$ @BillN Also, ρ has a standard use as a volume density, so seeing it in this relation could be confusing. Nope. The symbol $\rho$ is well defined in the text. $\endgroup$
    – Gert
    Jan 17, 2019 at 16:56
  • $\begingroup$ @BillN the usual derivation uses the linear density (mass per unit length) as in this answer. Your answer using the volume density makes the arbitrary assumptions that the cross section of the string is circular, and that the two strings have the same volume density That is not true for many stringed musical instruments, for example, where different strings may be made from different materials. For example "thin" piano "strings" are steel wire, but the thicker strings are a steel core surrounded by coiled copper, with different numbers of coiled layers for different groups of strings. $\endgroup$
    – alephzero
    Jan 17, 2019 at 19:05
  • $\begingroup$ @alephzero I understand all that. I am simply leveling the "playing field" to explain the OP question. What you have pointed out simply means that I can have a thick, low density string with the same fundamental as a thin high density string for a given tension and length. And yes, linear density is the usual presentation. How would you isolate the idea of thick versus thin? My answer points out specifically that thickness is not the only factor, but proceeds to a perspective in which it could be the only factor. Consider two solid steel strings, one 0.014" and another 0.028". $\endgroup$
    – Bill N
    Jan 17, 2019 at 22:21

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.