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As you can see in this wikipedia animation

enter image description here

Phase velocity (red) and group velocity (green) are described, and both show different "speed" in the Propagation direction but what about the transverse direction speed?

How do I calculate the rise speed of a cycle in transversal direction ?

(Look at the extremes of the picture the blue pixel going up and down, this is the vertical up-down speed what I want to calculate)

For example in a very simple wave like this

y(x,t) = A* sin(kx - w t), should I calculate derivative of y having x=constant?

What about in general? How to calculate the Rise speed?

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1 Answer 1

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Once you know the phase velocity $\omega$, you know that along the $x$ axis the red dot is described by $x(t) = \omega t$. Then you just apply the chain rule. The value of $y(x,t)$ at the red dot is a function of time $y(x(t),t)$. So $$ \frac{d}{dt}y(x(t),t) = \frac{\partial}{\partial x}y(x(t),t) \cdot \frac{d}{dt}x(t) + \frac{\partial}{\partial t} y(x(t),t) $$ which we more succinctly write as $$ \omega \partial_x y + \partial_t y $$

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To clarify, the position/velocity of a point on the wave can be modelled as SHM (simple harmonic motion), which is what Willie Wong is effectively doing. –  Noldorin Jul 22 '10 at 13:14
    
How can I compute the solution to a simple case, for example in wolfram alpha ? I think I would need to specify k=constant, w=constant –  Hernán Eche Jul 22 '10 at 13:20
    
k does not matter. –  Noldorin Jul 22 '10 at 13:54
    
This is the way in wolfram alpha to do that wD(ASin(kx - wt),x)+D(ASin(kx - w*t),t) –  Hernán Eche Jul 26 '10 at 15:41
1  
Then you are just looking at the vertical position change fixing the $x$-coordinate, which means that it is exactly given by just the partial derivative in the time-direction, $\partial_t$. –  Willie Wong Jul 26 '10 at 22:07

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