### Author Topic: 1-D Wave equation derivation  (Read 2730 times)

#### Zarak Mahmud ##### 1-D Wave equation derivation
« on: September 28, 2012, 01:27:42 PM »
Starting with

$$\begin{equation} \frac{\partial}{\partial x} \left[ T(x,t) \sin{\theta (x,t)} \right] = \rho (x) u_{tt} \end{equation}$$

where $\rho$ is the density and $T(x,t)$ is the tension force, we made the assumption that the vibrations are small, which gave us a linearized wave equation. I can see why some of the other assumptions (i.e. full flexibility, and no horizontal tension component) make sense, but I don't think I understand the insight behind this one.
« Last Edit: September 28, 2012, 02:04:08 PM by Zarak Mahmud »

#### Victor Ivrii ##### Re: 1-D Wave equation derivation
« Reply #1 on: September 28, 2012, 02:20:57 PM »
Starting with

\begin{equation} \frac{\partial}{\partial x} \left[ T(x,t) \sin{\theta (x,t)} \right] = \rho (x) u_{tt} \end{equation}

where $\rho$ is the density and $T(x,t)$ is the tension force, we made the assumption that the vibrations are small, which gave us a linearized wave equation. I can see why some of the other assumptions (i.e. full flexibility, and no horizontal tension component) make sense, but I don't think I understand the insight behind this one.

You mean that vibrations are small? Because usually they are. More general versions you find in some textbooks like
\begin{equation}
\rho u_{tt} = \Bigl(\frac{u_x}{\sqrt{1+u_x^2}}\Bigr)_x
\end{equation}
implicitly assume that displacement is only in the direction perpendicular to the string and that the density does not change--which is the case only for the small oscillations.