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)+\ldots$$=\sum _{{\text{odd }}n=1}^{\infty }{\frac {4}{n\pi }}\sin(n\pi x)$.

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 $kL_{0}=\pi$:

$\xi _{1}={\frac {\pi A^{2}}{8L_{0}}}\left(-\sin ^{2}(\omega t)+{\frac {c_{T}^{2}}{2c_{s}^{2}}}\right)\sin(2kX)$

${\displaystyle }$ ${\displaystyle }$ ${\displaystyle }$ ${\color {red}xx}<math>$ ${\displaystyle }$ ${\displaystyle }$ <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:

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