How much does string material elasticity affect tension at pitch for a stringed instrument?

Question

In introductory physics classes, we see the simple string oscillator equation, which can be used to estimate the string tension at pitch for a stringed instrument. However, the equation assumes an ideal string, meaning we don't consider string elasticity. Real stringed instruments are typically strung with steel, wound steel, nylon, or gut strings. How much does the elasticity of these materials affect the actual tension at pitch for these instrument strings?

For completeness, we could consider extremely elastic strings (rubber band guitars) too.

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(This question is inspired by comments on my answer here.)

Answer 1

If you consider the bending stiffness of the string you end up with the stiff string equation

$$T\frac{\partial^2 w}{\partial x^2} - EI\frac{\partial^4 w}{\partial x^4} = \mu \frac{\partial^2 w}{\partial t^2}\, ,$$

where $T$ is the tension, $E$ the Young modulus of the material, $I$ the second moment of area for the cross-section, and $\mu$ the linear mass density. This equation can be written in a non-dimensional form as

$$\frac{\partial^2 u}{\partial \xi^2} - \epsilon\frac{\partial^4 u}{\partial \xi^4} - \kappa^2 \frac{\partial^2 w}{\partial \tau^2} = 0\, ,$$

with $u = w/L$, $\tau=\omega t$, $\kappa^2 = L^2 \mu\omega^2/T = L^2 \omega^2/c^2$, $\epsilon = EI/(L^2 T)$, $L$ is the length of the string, $\omega$ a characteristic frequency of the system, and $c$ is the phase speed for the ideal string.

Notice that when $\epsilon$ is small this equation turns into the wave equation.

To find the vibration frequencies this problem is converted into the following eigenproblem

$$\frac{\partial^2 U}{\partial \xi^2} - \epsilon\frac{\partial^4 U}{\partial \xi^4} - \kappa^2 U = 0\, .$$

This problem has an analytical solution but the characteristic equation is not solvable analytically. You can use a perturbation method to get the following first-order approximation

$$\kappa^2 \approx n^2 \pi^2 + \epsilon n^4 \pi^4 \, .$$

For a G string in an electric guitar (196 Hz) the perturbation parameter $\epsilon$ is around $9.725 \times 10^{-6}$. So, in general, this parameter is small and the ideal string approximation is not that bad.

In terms of frequency, this translates to the following

$$f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}\sqrt{1 + \frac{n^2 \pi^2 E I}{L^2T}}\, .$$

References

• Zapata, N. G. (2021). Design of a fretboard using the stiff string equation. arXiv preprint arXiv:2106.13030. (Disclaimer: This is my own work).

• French, R. M. (2012). Technology of the Guitar. Springer Science & Business Media.

• Fletcher, N. H., & Rossing, T. D. (2012). The physics of musical instruments. Springer Science & Business Media.

Answer 2

In reality, transversal oscillations are not possible without change the length of the string. In the equilibrium position its length is the straight line between 2 fixed points. Any other path has bigger length, and that must happen during transverse oscillations.

Combining transverse and longitudinal waves:

$$\frac{\partial^2 u_y}{\partial t^2} = \frac{T}{\rho}\frac{\partial^2 u_y}{\partial x^2}$$ $$\frac{\partial^2 u_x}{\partial t^2} = \frac{E}{\rho}\frac{\partial^2 u_x}{\partial x^2} = \frac{1}{\rho}\frac{\partial T}{\partial x}$$

It is assumed for transverse oscillations that the tension in the string is pratically constant, what means that the longitudinal waves described by the second equation are negligible because $\frac{\partial T}{\partial x}\approx 0$.

If $E$ is too small (as for rubber string) that approximation is no more valid, and the first equation is not exactly a wave equation because $T = T(x)$, and $\frac{T}{\rho}$ is no longer a constant.