I came across this simple problem:´

A violin string has a density of $\rho=4.0\times 10^{-4}\space Kg\space m^{-1}$ and a length $L=0.33m$, and we want it to vibrate at a frequency $f=660Hz$, what is the tension on the string in that case?

The solution comes (partially) straightforwardly:Let $c$ be the speed of the wave and $F$ the tension. Then: $$F=\rho c^2\\c=\frac{\lambda}{T}\\F=\rho \lambda^2f^2$$

Now my problem is "What is the value of $\lambda$, the wavelength?" I could easily say that if the string is in the fundamental mode (the scenario where both ends are fixed) then $\lambda=L/2$, but then it gets me thinking that I can write the normal mode of vibration as a function of the tension of the string, and several questions arise: $$\lambda_n=\frac{2L}{n}\\n=\sqrt{\frac{\rho4L^2f^2}{F}}$$

1)If $\lambda$ can only take discrete values how come I can define it as a function of a continuous variable,$F$?

2) Does it mean that a string can only take discrete values of values of modes for a fixed tension value? But if that tension is a continuous variable then the string still has all the same modes but different normal frequencies, meaning that a string in a music instrument can take a continuous spectrum of normal frequencies given that the string is adjusted accordingly?

3) When a string of a violin is plucked it is vibrating at several modes all at once, and we know that each mode has it's own ressonant frequency, but it is way to hard to pluck a string just the right way to get that mode, then, is that why musicians stretch their strings? So as to get the desired mode of vibtration?

4) Usually we link an external oscilating force to get the desired ressonance of a string, but can the tension of a string of a music instrument "produce" a ressonance?


closed as too broad by David Z Dec 18 '18 at 10:56

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  • $\begingroup$ What you are missing is the effect of boundary conditions. That is what makes the modes discrete. The fact that the fundamental is related to a set of continuous variables does not contradict the discreteness of modes. And what does that have to do with "stretching"? $\endgroup$ – ggcg Dec 19 '18 at 12:54
  • $\begingroup$ The boundary conditions are that both ends are fixed (as stated), and the question that stretching gives is that when a string is stretched the tension in it is greater, hence ( And I admit that defining $n$ like that was not smart) something has to change in order to give the discretness of the modes of vibration $\endgroup$ – Bidon Dec 19 '18 at 13:28
  • 1
    $\begingroup$ I think you wording of the question is confusing. No, nothing has to change in order to give the discreteness of the modes. The modes are related to the fundamental by fn = n*f1, n = 1, 2, 3, 4.... This relationship is due to the B.C. and f1 is determined by the tension, mass density, etc. Change f1 and everything else changes with it. $\endgroup$ – ggcg Dec 19 '18 at 13:38

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