User:Egm6322.s09.three.liu/final report

=Wave Equation=

The Derivation of Wave Equation
Consider an infinite small segment of an elastic string shown in the picture, a small amplitude vibration of this string conform to the wave equation.



In the picture,

$$u(x,t)$$ is the vertical displacement of the string from the $$x$$ axis.

$$\theta(x,t$$) is the angle between the string and a horizontal line.

$$T(x,t)$$ is the tension in the string.

$$\rho(x)$$ is the mass density of the string.

$$F(x,t)$$ is the external force on the string.

Applying Newton's law in the vertical direction:

$$\rho(x) \sqrt{\Delta x^2+\Delta u^2} \frac {\partial^2 u}{\partial t^2}(x,t)=T(x+\Delta x,t)sin\theta(x+\Delta x,t)-T(x,t)sin\theta(x,t)+F(x,t)\Delta x$$

Divide $$\Delta x$$ on both RHS and LHS and take the limit as $$\Delta x \rightarrow 0$$ gives:

$$\rho(x)\sqrt{1+(\frac {\partial u}{\partial x})^2}\frac {\partial^2 u}{\partial t^2}(x,t)$$

$$=\frac {\partial}{\partial x}[T(x,t)sin\theta(x,t)]+F(x,t)$$

$$=\frac {\partial T}{\partial x}(x,t)sin\theta(x,t)+T(x,t)cos\theta(x,t)\frac{\partial \theta}{\partial x}{x,t}+F(x,t)$$ ...(1)

From the geometric relation in the picture,

$$sin\theta(x,t)=\frac{\frac{\partial u}{\partial x}(x,t)}{\sqrt{1+(\frac{\partial u}{\partial x}(x,t))^2}}$$

$$cos\theta(x,t)=\frac{1}{\sqrt{1+(\frac{\partial u}{\partial x}(x,t))^2}}$$

$$\theta(x,t)=tan^{-1}\frac{\partial u}{\partial x}(x,t)$$

$$ \frac{\partial \theta}{\partial x}(x,t)=\frac{\frac{\partial^2 u}{\partial x^2}(x,t)}{1+(\frac{\partial u}{\partial x}(x,t))^2}$$

Because $$\left | \theta(x,t) \right \vert<<1$$,

$$\sqrt{1+(\frac{\partial u}{\partial x})^2}\approx 1$$

$$ sin\theta(x,t)\approx\frac{\partial u}{\partial x}(x,t)$$

$$cos\theta(x,t)\approx1$$

$$\frac{\partial \theta}{\partial x}(x,t)\approx\frac{\partial^2 u}{\partial x^2}(x,t)$$

Substituting these into equation (1) gives

$$\rho(x) \frac{\partial^2 u}{\partial t^2}(x,t)=\frac{\partial T}{\partial x}(x,t)\frac{\partial u}{\partial x}(x,t)+T(x,t)\frac{\partial^2 u}{\partial x^2}+F(x,t)$$

Assume that the string only has transverse vibration, which means the horizontal force on it is zero.

So $$T(x+\Delta x,t)cos\theta(x+\Delta x,t)-T(x,t)cos\theta(x,t)=0$$

As $$\Delta x \rightarrow 0$$

$$\frac{\partial}{\partial x}[T(x,t)cos\theta(x,t)]=0$$

For small vertical vibration,$$cos\theta \rightarrow 1$$, and$$\frac{\partial T}{\partial x}(x,t)\rightarrow 0$$

Therefore, Equation (1) can be further simplified to:

$$\rho(x)\frac{\partial^2 u}{\partial t^2}(x,t)=T(t)\frac{\partial^2 u}{\partial x^2}(x,t)+F(x,t)$$

Assume the density is constant and no external forces, the second order 1-D wave equation finally goes to

$$\frac{\partial^2 u}{\partial t^2}(x,t)=c^2 \frac{\partial^2 u}{\partial x^2}(x,t), c=\sqrt{\frac{T}{\rho}}$$

In water waves, $$c$$ is the phase speed.