Guitar strings and fret positions ed

String description ed

Setup ed

We describe small movements of a string in a single plane by the function \(y(x)\)

Stretching ed

Our model ignores longitudinal waves! The string can only be stretched along the y-axis if \(y' \neq 0\):

Looking at a small piece \(\Delta x\) and stretching to \(\Delta x + \delta x\), by Hooke's law, we need the force \(F = k \, \delta x\) with \(k = E \, A\). Here \(E\) is Young's modulus, a material constant and about \(100GPa\) for steel. \(A=\pi r^2\) is the cross section area. We will also add \(F_0\), the force/tension used to tune the string.

Since string movement is small, we approximate \(\delta x\) by\[ \delta x = \Delta x \left(\sqrt{1 + y'^2} - 1\right) \approx \frac{\Delta x \, y'^2}{2} \]and the energy in this piece is\[ \Delta E_{\mathrm{stretch}} = \underbrace{\frac{E\,A}{8} \Delta x \, y'^4}_{\mathrm{ignore}} + F_0 \frac{\Delta x \, y'^2}{2} \]

Bending ed

...

String motion ed

Action ed

inspired by https://en.wikipedia.org/wiki/Euler–Bernoulli_beam_theory

\[ S = \int dt \int_0^L dx \left[ \frac{\mu}{2} (\dot y)^2 - \frac{E \, I}{2} (y'')^2 - \frac{F_0}{2} (y')^2 \right] = \iint \mathcal{L} \]

with \(E\) the Young modulus (\(\approx 100 GPa\)), \(\mu = \rho \, A = \rho \, \pi \, r^2\) the mass per unit length, \(I = \frac{\pi}{2} r^4\) the second moment of area, \(F_0 \approx 80N\) the initial tension/force on the string.

Differential equation ed

Euler-Lagrange\[ 0 = \frac{\partial \mathcal L}{\partial y} - \frac{\partial}{\partial t}\frac{\partial \mathcal L}{\partial \dot y} - \frac{\partial}{\partial x}\frac{\partial \mathcal L}{\partial y'} + \frac{\partial^2}{\partial x^2}\frac{\partial \mathcal L}{\partial y''} \]leads to\[ 0 = - \mu \, \ddot y + F_0 \, y'' - E \, I \, y'''' \]

We're looking for eigen-frequencies, using the Ansatz\[ y(x,t) = \hat y(x) \, \cos \omega t \]Now, the \(t\)-dependency splits off, leaving\[ 0 = \mu \, \omega^2 \, \hat y + F_0 \, \hat y'' - E\,I\,\hat y'''' \]Linear dgl., bla bla bla, insert \(e^{\beta x}\), solve for \(\beta^2\) gives\[ \beta^2 = \frac{F_0 \pm \sqrt{{F_0}^2 + 4 \mu\,\omega^2\,E\,I}}{2 E\,I} \]with \({\beta_+}^2 \gt 0 \Rightarrow \beta_+ \in \mathbb R\) and \({\beta_-}^2 \lt 0 \Rightarrow \beta_- \in i \mathbb R\). The 4 basic solutions correspond to\[ \cosh(\beta_+ x), \quad \sinh(\beta_+ x), \quad \cos(|\beta_-| x), \quad \sin(|\beta_-| x) \]

With the boundary conditions \(\hat y(0) = \hat y(L) = 0\), we'll pick the \(\sin\).

For the fundamental mode, we get \(\sin(|\beta_-| L) = 0 \Rightarrow |\beta_-| L = \pi\).

No bending term ed

\[ 0 = - \mu \, \ddot y + F_0 \, y'' \]Split time, insert \(e^{\beta x}\) gives\[ 0 = \mu \, \omega^2 + F_0 \, \beta^2, \quad \Rightarrow \quad \beta = \pm i \sqrt{\frac{\mu \, \omega^2}{F_0}} \]And the modes for \( \hat y = \sin(|\beta| x) \) set\[ |\beta_n| L = \sqrt{\frac{\mu \, \omega_n^2}{F_0}} L = n \, \pi \quad \Rightarrow \quad \omega_n = \frac{n \, \pi}{L} \sqrt{\frac{F_0}{\mu}} \]