# Finding the velocity of a wave

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1. Jan 14, 2016

### H Smith 94

I am currently studying a course on waves, which has a real ambiguity in the lecture notes. Essentially, I don't know how the professor got from equation \ref{eq:surf_x-y} to equations \ref{eq:vel_u} and \ref{eq:vel_w}. I have tried to work backwards to find a method but am not sure of its validity.

Please note that this is not a homework excercise, so full answers are appreciated.

Given formulas
The step is from the surface height of a 1-dimensional wave, given by $$\label{eq:surf_x-y}\eta(x,t) = A \cos(kx - \omega t)$$ to a velocity field $$\mathbf{v}(x,z,t) = (u(x,z,t),w(x,z,t)),$$ where $$\label{eq:vel_u}u(x,z,t) = A \omega \frac{\cosh(k(H+z))}{\sinh(k H)}\, \cos(kx - \omega t)$$ and $$\label{eq:vel_w}w(x,z,t) = A \omega \frac{\sinh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t).$$

My workings
Given this as true, one finds a position function $$\label{eq:pos_r}\phi(x,z,t) = \frac{A \omega}{k} \frac{\cosh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t)$$ from the integration $$\phi(x,z,t) = \int \mathbf{v}(x,z,t) \cdot\mathrm{d}\mathbf{r},$$ from which follows the assumption that we have multiplied equation \ref{eq:surf_x-y} by a $z$-dependent factor $$f(z) = \frac{\omega}{k}\,\frac{\cosh(k(H+z))}{\sinh(k H)}.$$

'Fudged method'
Now, the only way I can see of getting from \ref{eq:pos_phi} to $\phi(x,z,t)=\eta(x,t)\,f(z)$ is using the following method, which finds that:

$$\label{eq:vel_vec}\mathbf{v}(x,z,t) = \frac{1}{k}\,\frac{\partial}{\partial t}\,\nabla \phi(x,z,t) = \left(\begin{array}{c}\frac{\partial}{\partial t}\frac{\partial \phi}{\partial x} \\ \frac{\partial}{\partial t}\frac{\partial \phi}{\partial z}\end{array}\right).$$

Attempt at physical justification
I'm trying to justify this in physical terms by using the relation for phase speed $c$, which states that $c = \frac{\omega}{k}$, meaning we can infer from equation \ref{eq:vel_vec} that if $$\mathbf{v}(x,z,t) = \frac{\mathbf{\omega}}{k}$$ and so $$\mathbf{\omega} = \frac{\partial}{\partial t}\,\nabla \phi(x,z,t).$$

I am taking this to mean we have a scalar function $\phi(x,z,t)$, whose gradient gives the position $\mathbf{r}(x,y,t)=(r_x(t),r_y(t))$ of particles in the wave, meaning its rate of change would give us the velocity in the respective directions. If this is true, what is the physical meaning of $\phi$?

2. Jan 15, 2016

### nasu

It may help to formulate the problem in more detail. Maybe the source, too.
If it's 1-D wave what is the role of z coordinate? Is it a transverse wave?

3. Jan 15, 2016

### H Smith 94

I didn't want to put people off with too long a post, but you're right, I should provide more information. The textbook is available for free online, and it's equations (1.01) and (1.10-11) that I'm having problems with.

I don't understand either how the $z$ co-ordinate comes into play: my assumption was that $\eta$ represented vertcal surface height, which one would assume is the same as $z$. I believe it has something to do with phase space?

It's a transverse wave, yes.

I'm going to ask for more information after the lecture today, so hopefully that'll clear things up.