弦线上的波动方程推导

弦线上的波动方程推导

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1. 弦线上波动方程问题的四个基本假设

. 横震动
. 微小震动，既满足：
$$u(x+\Delta x,t)-u(x,t)\ll \Delta x$$

. 弦是柔软的假设，即张力沿着切线方向
. 弦是均匀的假设，密度均匀为 $\rho$

2. 一维非齐次波动方程推导

通过上图，利用牛顿第二定律可得，沿着x方向和垂直方向受力平衡方程为：
$$\{\begin{array}{l}{T}_2\cos {\alpha }_2-T_1\cos {\alpha }_1=0\\ T_2\sin {\alpha }_2-T_1\sin {\alpha }_1+F\mathrm{d}x=\rho \mathrm{d}x\frac{{\partial }^{2}u(x,t)}{\partial t^2}\end{array}\text{\hspace{1em}(1.1)}$$
下面利用四个假设条件进行化简上式。由于是微小震动，根据假设条件有：
$$\begin{array}{rl}u(x+\Delta x,t)-u(x,t)& \ll \Delta x\Rightarrow \\ \frac{u(x+\Delta x,t)-u(x,t)}{\Delta x}& \ll 1\stackrel{\Delta x\to 0}{\Rightarrow }\\ \underset{\Delta x}{\lim }\frac{u(x+\Delta x,t)-u(x,t)}{\Delta x}& =\frac{\partial u(x,t)}{\partial x}\ll 1\end{array}\text{\hspace{1em}(1.2)}$$
由于是微小横运动假设，则1.1式中三角函数有：
$$\begin{array}{rl}\sin {\alpha }_1\approx \tan {\alpha }_1& =\frac{u(x+\Delta x)-u(x,t)}{\Delta x}\stackrel{\Delta x\Rightarrow 0}{\Rightarrow }\\ 由式（1.2）\tan {\alpha }_1& =\frac{\partial u(x,t)}{\partial x}\end{array}\text{\hspace{1em}(1.3)}$$
$$\sin {\alpha }_2\approx \tan {\alpha }_2=\frac{\partial u(x+dx,t)}{\partial x}\text{\hspace{1em}(1.4)}$$
$$\begin{array}{rl}\cos {\alpha }_1& =\frac{\Delta x}{\sqrt{(\Delta x{)}^2+[u(x+\Delta x)-u(x,t){]}^2}}\\ & =\frac{1}{\sqrt{1+{\left[\frac{u(x+\Delta x,t)-u(x,t)}{\Delta x}\right]}^2}}\\ 由式(1.2)& =\frac{1}{\sqrt{1+{\left[\frac{\partial u(x,t)}{\partial x}\right]}^2}}\approx 1\end{array}\text{\hspace{1em}(1.5)}$$
同理有：
$$\cos {\alpha }_2=\frac{1}{\sqrt{1+{\left[\frac{\partial u(x+dx,t)}{\partial x}\right]}^2}}\approx 1\text{\hspace{1em}(1.6)}$$
综上所述，有：
$$\{\begin{array}{l}\sin {\alpha }_1\approx \tan {\alpha }_1=\frac{\partial u(x,t)}{\partial x}\\ \sin {\alpha }_2\approx \tan {\alpha }_2=\frac{\partial u(x+dx,t)}{\partial x}\\ \cos {\alpha }_1=\frac{1}{\sqrt{1+{\left[\frac{\partial u(x,t)}{\partial x}\right]}^2}}\approx 1\\ \cos {\alpha }_2=\frac{1}{\sqrt{1+{\left[\frac{\partial u(x+dx,t)}{\partial x}\right]}^2}}\approx 1\end{array}\text{\hspace{1em}(1.7)}$$
上式带入式(1.1)可得：
$$\{\begin{array}{l}{T}_2-T_1=0\\ T_2\frac{\partial u(x+dx,t)}{\partial x}-T_1\frac{\partial u(x,t)}{\partial x}+F(x,t)\mathrm{d}x=\rho \mathrm{d}x\frac{{\partial }^{2}u(x,t)}{\partial t^2}\end{array}\text{\hspace{1em}(1.8)}$$
即：
$$\begin{gathered} T_2=T_1=T \\ \Downarrow \\ T\frac{\left[\frac{\partial u(x+dx,t)}{\partial x}-\frac{\partial u(x,t)}{\partial x}\right]}{dx}dx+F(x,t)dx=\rho \mathrm{d}x\frac{{\partial }^{2}u(x,t)}{\partial t^2}\Rightarrow \\ T\frac{{\partial }^{2}u(x,t)}{\partial x^2}+F(x,t)=\rho \frac{{\partial }^{2}u(x,t)}{\partial t^2} \\ 令：c^2=\frac{T}{\rho },f(x,t)=\frac{F(x,t)}{\rho } \\ \Downarrow \\ \frac{{\partial }^{2}u(x,t)}{\partial t^2}=c^2\frac{{\partial }^{2}u(x,t)}{\partial x^2}+f(x,t)\text{\hspace{1em}(1.9)} \end{gathered}$$
式(1.9)为非齐次一维波动方程，分析式中c的量纲：
$$c^2=\frac{T}{\rho }\Rightarrow \frac{nt}{kg\setminus m}=\frac{kg\cdot m\backslash s^2}{kg\backslash m}=(\frac{m}{s}{)}^2$$
由上式可见，c的量纲为速度，即物理意义是波速，其值为波场除以周期（时间）：
$$c=\lambda \backslash T$$

3. 一维齐次波动方程推导

假设弦忽略重力$F(x,t)$作用，则平衡方程(1.2)可写成：
$$\{\begin{array}{l}{T}_2\cos {\alpha }_2-T_1\cos {\alpha }_1=0\\ T_2\sin {\alpha }_2-T_1\sin {\alpha }_1=\rho \Delta x\frac{{\partial }^{2}u(x,t)}{\partial t^2}\end{array}\text{\hspace{1em}(1.10)}$$
上式第一个式子有：
$$T_2\cos {\alpha }_2=T_1\cos {\alpha }_1=T\text{\hspace{1em}(1.11)}$$

用式(1.11)的三项，分别除式(1.10)第二个式子的左右两端有：
$$\begin{gathered} \frac{T_2\sin {\alpha }_2}{T_2\cos {\alpha }_2}-\frac{T_2\sin {\alpha }_1}{T_2\cos {\alpha }_1}=\frac{\rho }{T}\frac{{\partial }^{2}u(x,t)}{\partial t^2}\Delta x\Rightarrow \\ \tan {\alpha }_2-\tan {\alpha }_1=\frac{\rho }{T}\frac{{\partial }^{2}u(x,t)}{\partial t^2}\Delta x \\ \because 式(1.3)\Downarrow \\ \frac{\left[\frac{\partial u(x+\Delta x,t)}{\partial x}-\frac{\partial u(x,t)}{\partial x}\right]}{\Delta x}=\frac{\rho }{T}\frac{{\partial }^{2}u(x,t)}{\partial t^2} \\ 上式两部当\Delta x\to 时取极限\Downarrow \\ \underset{\Delta x\to 0}{\lim }\frac{\left[\frac{\partial u(x+\Delta x,t)}{\partial x}-\frac{\partial u(x,t)}{\partial x}\right]}{\Delta x}=\underset{\Delta x\Rightarrow 0}{\lim }\frac{\rho }{T}\frac{{\partial }^{2}u(x,t)}{\partial t^2}\Rightarrow \\ \frac{{\partial }^{2}u(x,t)}{\partial x^2}=\frac{1}{c^2}\frac{{\partial }^{2}u(x,t)}{\partial t^2}(c^2=\frac{T}{\rho }) \end{gathered}$$