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# The wave equation from modeling a vibrating string.
The 1-D wave equation is probably one of the most important partial differential equation in mathematics and physics, it states
$$
u_{tt} = c^2 u_{xx}
$$
for some suitable function $u=u(x,t)$. The convention of $c^2$ is just to emphasize a positive proportional constant.
This equation arises in many "wave like" modeling problems, and we shall focus on the vibrating string example in this article.
## Modeling a piece of vibrating string.
Suppose you have a piece of string of uniform mass density per unit length $\rho$, whose shape is given by some function $u(x,t)$, where $u$ denotes the height of the string at some position $x$ and time $t$, namely the $y$-position of the string.
One assumption is that at each point of the string it is constrained to have only vertical motion.
We seek a differential equation governing this string. My approach here is the same as [[0 inbox/What the shape dangling rope|the derivation for the shape of a hanging rope]], but I will recap them here.
For this string with shape given by $u(x,t)$, let us fix at a time $t$, the tiny segment of string from $x$ to $x+\Delta x$.
This tiny piece is being pulled to the left and right by tension forces from the rest of the string. Let us call the left tension force as $\vec T_L$, pulling the tiny segment to the left at position $x$; and write the right tension force as $\vec T_R$, pulling the tiny segment to the right at position $x+\Delta x$.
Now, using just the geometry, it is reasonable that the tension forces are in the direction tangent to the curve at those points.
So $\vec T_L$ is parallel to the vector $\langle 1, u_x(x,t) \rangle$, and $\vec T_R$ is parallel to the vector $\langle1, u_x(x+\Delta x,t)\rangle$.
Using our directionalities, we can write
$\vec T_L = \langle -k, -ku_x(x,t) \rangle$ for some positive $k > 0$, and
$\vec T_R = \langle k', k' u_x(x+\Delta x,t)\rangle$, for some positive $k' > 0$.
Since we assume there is no horizontal motion (only vertical), we must have $-k+k'= 0$, namely $k=k'$.
So the net force on this tiny segment of string has only $y$-component:
$$
k(u_x(x+\Delta x,t)-u_x(x,t))
$$
Now by Newton's law, that net force is given by mass times acceleration, we have roughly at position $x$,
$$
m u_{tt}(x,t) = k(u_x(x+\Delta x,t)-u_x(x,t))
$$
where $m$ is the mass of this tiny piece of string.
By Pythagorean theorem, this tiny piece of string has length $\sqrt{(\Delta x)^2+(\Delta u)^2}$, so mass is $m=\rho \sqrt{(\Delta x)^2+(\Delta u)^2}$ . Whence we get
$$
\rho \sqrt{(\Delta x)^2+(\Delta u)^2} \cdot u_{tt}(x,t) = k(u_x(x+\Delta x,t)-u_x(x,t))
$$
Dividing by $\Delta x$ we get
$$
\rho \sqrt{1+\left( \frac{\Delta u}{\Delta x} \right)^2} \cdot u_{tt}(x,t) = k \frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x}
$$
So in the limit of $\Delta x \to 0$, we get
$$
\rho \sqrt{1+\left(u_x \right)^2} \cdot u_{tt} = k u_{xx}
$$
Now, if we **further assume the amplitude of the wave is small**, then the slope $|u_x|\ll 1$, which we get the familiar wave equation:
$$
\rho u_{tt}=ku_{xx}
$$
////
## Post-script.
If one takes the density $\rho$ to be constant in horizontal distance (and not actual arclength), then the mass $m$ of the segment of the string from $x$ to $x+\Delta x$ is just $\rho\Delta x$. In this case we directly get the "wave equation".