How Does v^2 in the Wave Equation Represent Wave Speed?

[amiras]

The wave on the string could be described with wave equation.

Wave equation has a factor v^2 = Tension/linear density.

It has dimensions of speed, but from where exactly does it follow that this is actually speed of propagation of the wave?

 

[mfb]

v^2 has dimensions of speed squared. You can calculate the forces in the string based on the deflection in the string, and get this formula as a result. I'm sure there are textbooks where this derivation is done.

 

[HallsofIvy]

That can be interpreted as either (or both) of two questions:
1) How do we know that
$$\dfrac{\partial^2 y}{\partial x^2}= \frac{1}{c^2}\dfrac{\partial^2y}{\partial t^2}$$
(the "wave equation) does describe wave motion with wave speed "c"?

2) How do we show that the movement of a string, under tension T and with linear density $\delta$ satisfies the wave equation with $c= T/\delta$?

To answer the first one, suppose that f(x, t) satisfies the wave equation. Let g(x, t)= f(x- ct_0, t- t_0). Show that g(x, t) also satisfies the wave equation. Of course, g(x, t) is just f(x, t) with time shifted by $t_0$ and distance shifted by $c t_0$ so that the difference in distances is $ct_0$ for a change in time $t_0$ so that the speed is $ct_0/t_0= c$.

Before answering the second, I have to admit it isn't exactly true! It is an approximation but a very good approximation for small waves. If you were to pull on the string or wire really hard, you might well permanently stretch or even break it!

Imagine a small part of the string between $(x_0, y(x_0))$ and $x_0+ \Delta x, y(x_0+\Delta x))$. Let the angle the string makes at $x_0$ be $\theta_{x_0}$. With tension, T, the upward force at $x_0$ is $T sin(\theta_{x_0}$. Let the angle at $x_0+ \Delta x$ be $\theta_{x_0+ \Delta x}$. The upward force at $x_0+ \Delta x$ is $T sin(\theta_{x_0+ \Delta x}$ so the net force is $T(sin(\theta_{x_0+ \Delta x})- sin(\theta_{x_0}))$. For small angles sine is approximately tangent so we can approximate that by $T(tan(\theta_{x_0+ \Delta x}- tan(\theta_{x_0}))$.

The tangent of the angle of a curve, at any point, is the derivative with respect to x so can write that as
$$T(\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0}$$

If that section of string has length s and density $\delta$ then it has mass $\delta \Delta s$. For small angles, we can approximate $\Delta s$ by $\Delta x$. "Mass times acceleration" is
$$\delta \Delta x \dfrac{\partial y^2}{\partial t^2}$$
so that "force equals mass times acceration" becomes
$$T(\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0})= \delta \Delta x \dfrac{\partial^2 y}{\partial t^2}$$

Dividing both sides by $T \Delta x$, that becomes
$$\dfrac{\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0}}{\Delta x}= \dfrac{\delta}{T}\dfrac{\partial^2 y}{\partial t^2}$$

Now we can recognize that fraction on the left as a "difference quotient" which, in the limit as $\Delta x$ goes to 0, becomes the derivative with respect to x. Since the function in the difference quotient is the first derivative of y with respect to x, the limit gives the second derivative:
$$\dfrac{\partial^2y}{\partial x^2}= \dfrac{\delta}{T}\dfrac{\partial^2 y}{\partial t^2}$$

the wave equation with $1/c^2= \delta/T$ so that $c^2= T/\delta$.

 

[The_Duck]

The wave on the string could be described with wave equation.

Wave equation has a factor v^2 = Tension/linear density.

It has dimensions of speed, but from where exactly does it follow that this is actually speed of propagation of the wave?

Convince yourself that for an arbitrary function $f(x)$ the function $y(x, t) = f(x - vt)$ solves the wave equation. Then convince yourself that this solution describes a wave whose shape is given by $f$ and which propagates in the positive $x$ direction at a speed $v$.