# Physics/A/String vibration/Nonlinear

Define $\kappa _{T}=(1-a)\kappa _{s}\leq \kappa _{s}$
$m{\ddot {\xi }}=\kappa _{s}\xi ^{\prime \prime }+a\kappa _{s}\eta ^{\prime }\eta ^{\prime \prime }$, 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 ${\mathcal {L}}=m\partial ^{2}/\partial t^{2}-\kappa _{s}\partial ^{2}/\partial X^{2}$
Second order differential equation with one variable: https://openstax.org/books/calculus-volume-3/pages/7-2-nonhomogeneous-linear-equations
${\mathcal {L}}\xi =-k^{3}A^{2}\sin(kx)\cos(kx)\sin ^{2}(\omega t)$
$\xi =\xi _{p}(X,t)+\xi _{h}(X,t)$ where $\xi _{h}$ is the solution to the homogeneous equation, i.e., solution to ${\mathcal {L}}\xi _{h}=0$
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 $\sin ^{2}(\omega t)={\frac {1-\cos(2\omega t)}{2}}$
${\begin{aligned}{\mathcal {L}}\xi &=-a\kappa _{s}k^{3}A^{2}\sin(2kX)\left({\frac {1-\cos(2\omega t)}{4}}\right)\\&=-{\frac {a\kappa _{s}k^{3}A^{2}}{4}}\sin(2kX)+{\frac {a\kappa _{s}k^{3}A^{2}}{4}}\sin(2kX)\cos(2\omega t)\end{aligned}}$
To find a particular solution, $\xi _{p},$ to (?) we first consider two different inhomogeneous equations:
${\mathcal {L}}\xi _{1}=-{\frac {a\kappa _{s}k^{3}A^{2}}{4}}\sin(2kX)$
${\mathcal {L}}\xi _{2}={\frac {a\kappa _{s}k^{3}A^{2}}{4}}\sin(2kX)\cos(2\omega t)$

## 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 ${\mathcal {L}}\xi _{1}=4k^{2}\kappa _{s}\xi _{1}$, and: $\xi _{1}=-{\frac {a\kappa _{s}kA^{2}}{16\kappa _{T}}}\sin(2kX)$ => ${\color {red}\xi _{1}=-{\frac {kA^{2}}{16}}\left(1-{\frac {\kappa _{T}}{\kappa _{s}}}\right)\sin(2kX)}$
If $\xi _{2}$ is proportional to $\sin(2kX)\cos(2\omega t)$, then ${\mathcal {L}}\xi _{2}=\left(-4m\omega ^{2}+4k^{2}\kappa _{s}\right)\xi _{2}$ and:$\xi _{2}={\frac {a\kappa _{s}}{16}}{\frac {k^{3}A^{2}}{k^{2}\kappa _{s}-m\omega ^{2}}}\sin(2kX)\cos(2\omega t)$ =$\xi _{2}={\frac {a\kappa _{s}kA^{2}}{16}}{\frac {1}{\kappa _{s}-m\omega ^{2}/k^{2}}}\sin(2kX)\cos(2\omega t)$ =>$\xi _{2}={\frac {a\kappa _{s}kA^{2}}{16}}{\frac {1}{\kappa _{s}-\kappa _{T}}}\sin(2kX)\cos(2\omega t)$ =>$\xi _{2}={\frac {kA^{2}}{16}}\sin(2kX)\cos(2\omega t)$. Now use $\cos(2\omega t)=1-2\sin ^{2}\omega t)$.
${\color {red}\xi _{1}={\frac {kA^{2}}{16}}\left(1-2\sin ^{2}\omega t\right)\sin(2kX)}$
By the linearity of the operator ${\mathcal {L}},$ we see that a particular solution to (?) is the sum of $\xi _{p}=\xi _{1}+\xi _{2}:$
$\xi _{1}={\frac {kA^{2}}{8}}\left(-\sin ^{2}(\omega t)+{\frac {\kappa _{T}}{2\kappa _{s}}}\right)\sin(2kX)$
In these units the speed of a $\left\{{\text{transverse, longitudinal}}\right\}$ wave is $\left\{c_{T}={\sqrt {\kappa _{T}/m}},\,c_{s}={\sqrt {\kappa _{s}/m}}\right\}$. 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 $kL_{0}=\pi$:
$\xi _{1}={\frac {\pi A^{2}}{8L_{0}}}\left(-\sin ^{2}(\omega t)+{\frac {c_{T}^{2}}{2c_{s}^{2}}}\right)\sin(2kX)$
$$ $$ $$ ${\color {red}xx}<math>$ $$ $$ <math></math>

## Other identities

wikipedia:special:permalink/1017302768
$\sin(2\theta )=2\sin \theta \cos \theta$ *  $\cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1=1-2\sin ^{2}\theta$ *  $\sin ^{2}\theta ={\frac {1-\cos(2\theta )}{2}}$ *  $\cos ^{2}\theta ={\frac {1+\cos(2\theta )}{2}}$ *  $\sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$ *  $\cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$ *  $2\sin \theta \cos \varphi ={\sin(\theta +\varphi )+\sin(\theta -\varphi )}$ *  $2\cos \theta \sin \varphi ={\sin(\theta +\varphi )-\sin(\theta -\varphi )}$ *
After modifying an equation from Wikipedia:
${\begin{aligned}\sin \alpha \cos \beta &={\frac {e^{i\alpha }-e^{-i\alpha }}{2i}}\cdot {\frac {e^{i\beta }+e^{-i\beta }}{2}}\\&={\frac {1}{2}}\cdot {\frac {e^{i(\alpha +\beta )}+e^{i(\alpha -\beta )}-e^{i(-\alpha +\beta )}-e^{i(-\alpha -\beta )}}{2i}}\\&={\frac {1}{2}}{\bigg\underbrace {\frac {e^{i(\alpha +\beta )}-e^{-i(\alpha +\beta )}}{2i}} _{\sin(\alpha +\beta )}+\underbrace {\frac {e^{i(\alpha -\beta )}-e^{-i(\alpha -\beta )}}{2i}} _{\sin(\alpha -\beta )}{\bigg )}.\end{aligned}}$
* ↑ See David R Rowland 2011 Eur. J. Phys. 32 1475, equation 11