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Basic question on wave equation - need a reminder!

  1. Mar 10, 2015 #1
    Greetings all and new to the forum here.

    It's been many years and I've forgotten how to do it, and it should be a basic question, but assuming we have an equation Ex=E_0*cos(wt-kz), how do we translate to sine? I've seen it written sin(kz-wt) or sin(wt-kz), but I've just plainly forgotten how to get from cos(wt-kz) to sin nomenclature, and which dependance is right (kz-wt or wt-kz).

  2. jcsd
  3. Mar 10, 2015 #2


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    For starters, you can say:
    [itex]Cos(\omega t - k z) = Sin\big(\frac{\pi}{2}-(\omega t - k z)\big)[/itex]
    for the same reason that
    [itex]Cos(x) = Sin\big(\frac{\pi}{2}-x\big)[/itex]

    Then, for a wave, where the [itex]z[/itex]-coordinate of its peak increases with time (so the wave travels in the positive [itex]z[/itex] direction), the most popular convention is [itex](kz-\omega t)[/itex] (both work, but you need to stick to one for whatever problem you're doing).

    To see how this works, you can think that where [itex](kz-\omega t)=0[/itex], you have that [itex]Cos(kz-\omega t)=1[/itex], a peak.

    As [itex]t[/itex] increases, we have to have that [itex]z=\frac{\omega}{k}t[/itex] to be at the location where that peak is (i.e., where [itex](kz-\omega t)=0[/itex]).

    So with this convention, the [itex]z[/itex]-coordinate of that peak moves in the positive [itex]z[/itex]-direction with a velocity [itex]\frac{\omega}{k}[/itex]. This velocity is also known as the phase velocity of the wave.
    Last edited: Mar 10, 2015
  4. Mar 16, 2015 #3
    Thanks, that cleared it up!

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