📂Physics

Derivation of the One-Dimensional Wave Equation

Overview

The one-dimensional wave equation is as follows.

$$ \dfrac{\partial ^{2} f }{\partial x^{2}} = \dfrac{1}{v^{2}}\dfrac{\partial ^{2} f}{\partial t^{2}} $$

Here, $v$ represents the propagation speed of the wave.

Characteristics of Waves

Let’s assume there is a wave with a constant speed of $v$ as shown in Figure 1. Let the displacement of the point at $x$ at time $t$ be $f(x,t)$. Assuming the initial displacement of the string is $g(x)=f(x,0)$, after $t$ seconds, we want to know the displacement of the string. Since the speed is $v$, it is equivalent to having been translated by $vt$, as shown in Figure 2. Therefore,

$$ \begin{equation} f(x,t)=f(x-vt,0)=g(x-vt) \end{equation} $$

This equation tells us that the wave function is a function of only $x-vt$, which combines the two variables $x,\ t$. Therefore, $f_{1}$, $f_2$, and $f_{3}$ represent waves, but $f_{4}$ and $f_{5}$ do not.

$$ f_{1}=Ae^{-(x-vt)^{2}}, \quad f_2=A\sin\big( 2(x-vt) \big),\quad f_{3}=\dfrac{A}{(x-vt)^{2}-1} \\ f_{4}=Ae^{x(x-vt)},\quad f_{5}=A\cos(x) \cos(xvt) $$

Derivation

Method 1 ¹

By considering the motion of a tightly stretched string, we can derive the one-dimensional wave equation. When the string deviates from its equilibrium position, the force due to tension $T$ acting vertically on a segment of length $\Delta x$ can be represented as

$$ \Delta F=T\sin \theta^{\prime} - T\sin \theta $$

When $\theta$ is sufficiently small, since $\sin \theta \approx \tan \theta$, the above equation can be written as

$$ \Delta F \approx T(\tan \theta^{\prime} -\tan \theta) $$

As $\tan$ represents the gradient (derivative),

$$ \begin{equation} \begin{aligned} \Delta F \approx&\ T(\tan \theta^{\prime} -\tan \theta) \\ =&\ T \big[ f^{\prime}(x+\Delta x) -f^{\prime}(x) \big] \\ \approx&\ T\dfrac{\partial ^{2} f}{\partial x^{2}}\Delta x \quad \end{aligned} \end{equation} $$

Let’s denote the mass per unit length of the string as $\mu$. Then, according to Newton’s second law, $(F=ma)$,

$$ \begin{equation} \begin{aligned} \Delta F =&\ m\Delta a \\ =&\ m \dfrac{\partial^{2} f}{\partial t^{2}} \\ =&\ \mu \Delta x \dfrac{\partial ^{2} f}{\partial t^{2}} \quad \end{aligned} \end{equation} $$

By $(1)$ and $(2)$,

$$ \dfrac{\partial^{2} f}{\partial x^{2}}=\dfrac{\mu}{T} \dfrac{\partial^{2} f}{\partial t^{2}} $$

Substituting with $\sqrt{\frac{T}{\mu}}=v$,

$$ \frac{\partial^{2} f}{\partial x^{2}}=\dfrac{1}{v^{2}}\dfrac{\partial^{2} f}{\partial t^{2}} $$

This is referred to as the one-dimensional wave equation. The solution to this equation must satisfy the following form (as differentiation twice with respect to $t$ should yield a term of $v^{2}$).

$$ f(x,t)=g(x-vt) $$

Here, we can see that $v$, as discussed above, represents the propagation speed of the wave.

■