Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the velocity of a wave

  1. Jan 14, 2016 #1

    H Smith 94

    User Avatar
    Gold Member

    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 \begin{equation}\label{eq:surf_x-y}\eta(x,t) = A \cos(kx - \omega t)\end{equation} to a velocity field $$\mathbf{v}(x,z,t) = (u(x,z,t),w(x,z,t)),$$ where \begin{equation}\label{eq:vel_u}u(x,z,t) = A \omega \frac{\cosh(k(H+z))}{\sinh(k H)}\, \cos(kx - \omega t)\end{equation} and \begin{equation}\label{eq:vel_w}w(x,z,t) = A \omega \frac{\sinh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t).\end{equation}

    My workings
    Given this as true, one finds a position function \begin{equation}\label{eq:pos_r}\phi(x,z,t) = \frac{A \omega}{k} \frac{\cosh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t)\end{equation} 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:

    \begin{equation}\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).\end{equation}

    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 \begin{equation}\mathbf{\omega} = \frac{\partial}{\partial t}\,\nabla \phi(x,z,t).\end{equation}

    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. jcsd
  3. Jan 15, 2016 #2
    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?
  4. Jan 15, 2016 #3

    H Smith 94

    User Avatar
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Finding the velocity of a wave