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 \(dx\) and stretching to \(dx + \delta x\), 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/tensionat rest from tuning the string.
Since string movement is small, we approximate \(\delta x\) by\[ \delta x = \left(\sqrt{1 + (y')^2} - 1\right) \, dx \approx \frac{(y')^2}{2} \,dx \]and the energy per piece is\[ dE_{\mathrm{stretch}} = \underbrace{\frac{E\,A}{8} \, (y')^4 \, dx}_{\mathrm{ignore}} + F_0 \frac{(y')^2}{2}\,dx \]
Bending ed
Bending is more complicated. The curvature of the string is \(y''\) (as long as we only allow small movements). The resistance to bending is \(E\,I\). \(E\) is again Young's modulus (for most materials) and \(I=\frac{\pi}{4} r^4\) is the second moment of area.
The energy for a small piece is\[ dE_{\mathrm{bend}} = \frac{E\,I}{2} (y'')^2 \, dx \]
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\[ |\beta_n| L = \sqrt{\frac{\mu \, \omega_n^2}{F_0}} L = n \, \pi \quad \Rightarrow \quad \omega_n = \frac{n \, \pi}{L} \sqrt{\frac{F_0}{\mu}} \]