Physics/A/String vibration/Nonlinear

If you don't understand this animated gif, look at the next figure

The first four terms of the Fourier series of a square wave, $\psi (x)=\frac{4}{\pi }\sin (\pi x)$ $+\frac{4}{3\pi }\sin (3\pi x)$ $+\frac{4}{5\pi }\sin (5\pi x)$ $+\frac{4}{7\pi }\sin (7\pi x)+\dots$$=\sum _{\text{odd }n=1}^{\mathrm{\infty }}\frac{4}{n\pi }\sin (n\pi x)$.

Define ${\kappa }_T=(1-a){\kappa }_s\le {\kappa }_s$

$m\ddot{\xi }={\kappa }_s{\xi }^{\prime \prime }+a{\kappa }_s{\eta }^{\prime }{\eta }^{\prime \prime }$, and $m\ddot{\eta }={\kappa }_T{\eta }^{\prime \prime }$.

Transverse standing wave: $\omega /k=\sqrt{{\kappa }_T/m}$

$\eta =A\sin (kx)\sin (\omega t)$

${\eta }^{\prime }{\eta }^{\prime \prime }=-k^3{A}^2\sin (kx)\cos (kx){\sin }^2(\omega t)$

Define $ℒ=m{\partial }^2/\partial t^2-{\kappa }_s{\partial }^2/\partial X^2$

Second order differential equation with one variable: https://openstax.org/books/calculus-volume-3/pages/7-2-nonhomogeneous-linear-equations

$ℒ\xi =-k^3{A}^2\sin (kx)\cos (kx){\sin }^2(\omega t)$

$\xi ={\xi }_p(X,t)+{\xi }_h(X,t)$ where ${\xi }_h$ is the solution to the homogeneous equation, i.e., solution to $ℒ{\xi }_h=0$

Link to wikipedia:Fourier series?

• https://www.mathsisfun.com/calculus/fourier-series.html
• https://mathworld.wolfram.com/FourierSeries.html

Employ two identities:

$\sin (kX)\cos (kX)=\frac{\sin (2kX)}{2}$ and ${\sin }^2(\omega t)=\frac{1-\cos (2\omega t)}{2}$

$\begin{array}{cc}ℒ\xi & =-a{\kappa }_s{k}^3{A}^2\sin (2kX)\left(\frac{1-\cos (2\omega t)}{4}\right)\\ & =-\frac{a{\kappa }_s{k}^3{A}^2}{4}\sin (2kX)+\frac{a{\kappa }_s{k}^3{A}^2}{4}\sin (2kX)\cos (2\omega t)\end{array}$

To find a particular solution, ${\xi }_p,$ to (?) we first consider two different inhomogeneous equations:

$ℒ{\xi }_1=-\frac{a{\kappa }_s{k}^3{A}^2}{4}\sin (2kX)$

$ℒ{\xi }_2=\frac{a{\kappa }_s{k}^3{A}^2}{4}\sin (2kX)\cos (2\omega t)$

NOW

Recall ${\kappa }_T=(1-a){\kappa }_s$ => $a=\frac{{\kappa }_s-{\kappa }_T}{{\kappa }_s}$

If ${\xi }_1$ is proportional to $\sin (2kX)$, then $ℒ{\xi }_1=4k^2{\kappa }_s{\xi }_1$, and: ${\xi }_1=-\frac{a{\kappa }_{s}k{A}^2}{16{\kappa }_T}\sin (2kX)$ => ${\xi }_1=-\frac{k{A}^2}{16}(1-\frac{{\kappa }_T}{{\kappa }_s})\sin (2kX)$

If ${\xi }_2$ is proportional to $\sin (2kX)\cos (2\omega t)$, then $ℒ{\xi }_2=(-4m{\omega }^2+4k^2{\kappa }_s){\xi }_2$ and:${\xi }_2=\frac{a{\kappa }_s}{16}\frac{k^3{A}^2}{k^2{\kappa }_s-m{\omega }^2}\sin (2kX)\cos (2\omega t)$ =${\xi }_2=\frac{a{\kappa }_{s}k{A}^2}{16}\frac{1}{{\kappa }_s-m{\omega }^2/k^2}\sin (2kX)\cos (2\omega t)$ =>${\xi }_2=\frac{a{\kappa }_{s}k{A}^2}{16}\frac{1}{{\kappa }_s-{\kappa }_T}\sin (2kX)\cos (2\omega t)$ =>${\xi }_2=\frac{k{A}^2}{16}\sin (2kX)\cos (2\omega t)$. Now use $\cos (2\omega t)=1-2{\sin }^2\omega t)$.

${\xi }_1=\frac{k{A}^2}{16}(1-2{\sin }^2\omega t)\sin (2kX)$

By the linearity of the operator $ℒ,$ we see that a particular solution to (?) is the sum of ${\xi }_p={\xi }_1+{\xi }_2:$

${\xi }_1=\frac{k{A}^2}{8}(-{\sin }^2(\omega t)+\frac{{\kappa }_T}{2{\kappa }_s})\sin (2kX)$

In these units the speed of a $\left\{\text{transverse, longitudinal}\right\}$ wave is $\{c_T=\sqrt{{\kappa }_T/m},c_s=\sqrt{{\kappa }_s/m}\}$. This permits us to write an expression that does not depend on the choice of units. Relating the wavenumber of the lowest order mode to string length by $k{L}_0=\pi$:

${\xi }_1=\frac{\pi A^2}{8L_0}(-{\sin }^2(\omega t)+\frac{c_T^2}{2c_s^2})\sin (2kX)$

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Other identities

wikipedia:special:permalink/1017302768

$\sin (2\theta )=2\sin \theta \cos \theta$Template:Spaces*Template:Spaces $\cos (2\theta )={\cos }^2\theta -{\sin }^2\theta =2{\cos }^2\theta -1=1-2{\sin }^2\theta$Template:Spaces*Template:Spaces ${\sin }^2\theta =\frac{1-\cos (2\theta )}{2}$Template:Spaces*Template:Spaces ${\cos }^2\theta =\frac{1+\cos (2\theta )}{2}$Template:Spaces*Template:Spaces $\sin (\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$Template:Spaces*Template:Spaces $\cos (\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$Template:Spaces*Template:Spaces $2\sin \theta \cos \phi =\sin (\theta +\phi )+\sin (\theta -\phi )$Template:Spaces*Template:Spaces $2\cos \theta \sin \phi =\sin (\theta +\phi )-\sin (\theta -\phi )$Template:Spaces*Template:Spaces

After modifying an equation from Wikipedia:

$\begin{array}{cc}\sin \alpha \cos \beta & =\frac{e^{i\alpha }-e^{-i\alpha }}{2i}\cdot \frac{e^{i\beta }+e^{-i\beta }}{2}\\ & =\frac{1}{2}\cdot \frac{e^{i(\alpha +\beta )}+e^{i(\alpha -\beta )}-e^{i(-\alpha +\beta )}-e^{i(-\alpha -\beta )}}{2i}\\ & =\frac{1}{2}(\underset{\sin (\alpha +\beta )}{\underbrace{\frac{e^{i(\alpha +\beta )}-e^{-i(\alpha +\beta )}}{2i}}}+\underset{\sin (\alpha -\beta )}{\underbrace{\frac{e^{i(\alpha -\beta )}-e^{-i(\alpha -\beta )}}{2i}}}).\end{array}$