Physics/A/String vibration/Nonlinear



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: $$\omega/k=\sqrt{\kappa_T/m}$$

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

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

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

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

$$\xi = \xi_p(X,t) + \xi_h(X,t)$$ where $$\xi_h$$ is the solution to the homogeneous equation, i.e., solution to $$\mathcal L \xi_h=0$$

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

$$\begin{align} \mathcal L \xi &= -a\kappa_sk^3A^2\sin(2kX)\left(\frac{1-\cos(2\omega t)}{4}\right)\\ &=-\frac{a\kappa_sk^3A^2}{4}\sin(2kX) +\frac{a\kappa_sk^3A^2}{4}\sin(2kX)\cos(2\omega t) \end{align}$$

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

$$\mathcal L \xi_1 =-\frac{a\kappa_sk^3A^2}{4}\sin(2kX)$$

$$\mathcal L \xi_2 = \frac{a\kappa_sk^3A^2}{4}\sin(2kX)\cos(2\omega t)$$

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: $$ \xi_1=-\frac{a\kappa_skA^2}{16\kappa_T}\sin(2kX)$$ => $${\color{red} \xi_1=-\frac{kA^2}{16}\left(1-\frac{\kappa_T}{\kappa_s}\right)\sin(2kX)}$$

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: $$\xi_2=\frac{a\kappa_s}{16}\frac{k^3A^2}{k^2\kappa_s-m\omega^2}\sin(2kX)\cos(2\omega t)$$ =$$\xi_2=\frac{a\kappa_skA^2}{16}\frac{1}{\kappa_s-m\omega^2/k^2}\sin(2kX)\cos(2\omega t)$$ =>$$\xi_2=\frac{a\kappa_skA^2}{16}\frac{1}{\kappa_s-\kappa_T}\sin(2kX)\cos(2\omega t)$$ =>$$ \xi_2=\frac{kA^2}{16}\sin(2kX)\cos(2\omega t)$$. 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}

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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:

$$\begin{align} \sin \alpha \cos \beta &= \frac{e^{i\alpha}-e^{-i\alpha}}{2i} \cdot \frac{e^{i\beta}+e^{-i\beta}}{2} \\ &= \frac{1}{2}\cdot \frac{e^{i(\alpha+\beta)}+e^{i(\alpha-\beta)}-e^{i(-\alpha+\beta)}-e^{i(-\alpha-\beta)}}{2i} \\ &= \frac12 \bigg( \underbrace{ \frac{e^{i(\alpha+\beta)} - e^{-i(\alpha+\beta)}}{2i} } _{\sin(\alpha+\beta)} + \underbrace{ \frac{e^{i(\alpha-\beta)}  - e^{-i(\alpha-\beta)}}{2i} } _{\sin(\alpha-\beta)}  \bigg). \end{align}$$