# Why does the wave equation describe certain phenomena

• I
I have recently started learning about waves. We didn't really formally describe what a wave is, but instead started by looking at a concrete example namely harmonic sinusoidal waves in 1d.

We then introduced the wave equation in 1d and showed that the sinusoidal waves indeed satisfy this equation.

So to my current understanding a 1d wave is a phenomena described by a function which associates to each point in time a scalar displacement at each point on a line, that satisfies the wave equation. So far so good.

Now there is an example where a rod attached to a wall is pulled and we want to analyse the displacement at each point on the rod from its original position over time. Using hooks Law we are able to derive a relationship between the second partial of time and the second partial of space(of the displacement). We then plug this into the wave equation, which gives us an expression for the velocity at which this wave propagates through the rod.

My question is, why are we able to plug these values into the wave equation? What about this example tells us that these displacements will indeed satisfy the wave equation?

## Answers and Replies

Ibix
Science Advisor
2020 Award
Well, what should have happened is that you develop a differential equation describing the motion of the rod. Then you simply recognise that the particular combination of differentials you have got is one you've seen before - the wave equation. And you already know the solutions to that.

So it's not really plugging it in to the wave equation. It's recognising that your equation is the wave equation. And because you know how to read the wave speed off the wave equation, you can just read off the wave speed of your particular phenomenon.

I may have expressed myself poorly with "plugging into the wave equation". We do derive a differential equation that looks just like the wave equation except that instead of v^2 we have a different expression. Dont we have to make the assumption that this phenomena will satisfy the wave equation to be able to conclude that this expression must be equal to v^2?

Nugatory
Mentor
I may have expressed myself poorly with "plugging into the wave equation". We do derive a differential equation that looks just like the wave equation except that instead of v^2 we have a different expression. Dont we have to make the assumption that this phenomena will satisfy the wave equation to be able to conclude that this expression must be equal to v^2?
Can you provide a specific example of what's perplexing you? The wave equation relates the second derivative of a dynamical variable to the second derivative of time in a particular way. The sequence that we usually see is that we derive a differential equation from our knowledge of the dynamics of the system, look at it, and recognize it as the wave equation. Therefore this phenomenon obeys the wave equation, in a rather trivially obvious way.

Ibix
Science Advisor
2020 Award
Well, you've just derived the equation describing the system from physical considerations. So you don't assume anything - you know it satisfies that equation if you didn't make any mistakes.

What you do then is up to you. You don't need to notice that it's the wave equation - you can solve it from first principles. But why bother? The answers aren't going to be different whether you recognise the equation and look up the results or if you grind through the maths yourself.

Mathematically ##v^2## is just a constant. If you have a different constant ##A## then grind through the maths you'll get wave solutions with ##v=\sqrt A##. Or you can notice that if you write ##v=\sqrt A## then you've got the exact wave equation and you're done. Up to you.

My differential equation tells me that the second partial of time is equal to a constant times the second partial of space. How can I conclude from this that the constant is v^2 without referencing the wave equation?

Ibix
Science Advisor
2020 Award
By solving it. Naturally you will find wave-like solutions because it's the wave equation, but if you don't mind solving differential equations you don't need to acknowledge that.

Fibo112
Ooh, I see. So in the solution function the value at (x,t1) will be equal to the value at (x+a*t2,t1+t2) where a is the root of my constant.

I guess the wave equation wouldnt make much sense at all if this werent the case!