# Basic question on wave equation - need a reminder!

1. Mar 10, 2015

### VictorVictor5

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).

Thanks!
VV5

2. Mar 10, 2015

### jfizzix

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

Then, for a wave, where the $z$-coordinate of its peak increases with time (so the wave travels in the positive $z$ direction), the most popular convention is $(kz-\omega t)$ (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 $(kz-\omega t)=0$, you have that $Cos(kz-\omega t)=1$, a peak.

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

So with this convention, the $z$-coordinate of that peak moves in the positive $z$-direction with a velocity $\frac{\omega}{k}$. This velocity is also known as the phase velocity of the wave.

Last edited: Mar 10, 2015
3. Mar 16, 2015

### VictorVictor5

Thanks, that cleared it up!

VV5