# Wave equation and wave speed

1. Nov 6, 2013

### 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?

2. Nov 6, 2013

### Staff: Mentor

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.

3. Nov 6, 2013

### HallsofIvy

Staff Emeritus
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 $c^2= T/\delta$.

Last edited: Nov 6, 2013
4. Nov 6, 2013

### The_Duck

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 $v$.