OK, so you know how the normal formula is y=Asin(wt-kx) but now they've given an equation with cos. I need to find the propagation speed of the wave following this equation: y=2cospi(x-2t) Help

Just to be sure, the equation given to you is:

y = 2\cos(\pi(x-2t))

?

Remember that cos(x) is the same as sin(x+pi/2). So you could rewrite the equation:

y=2 \cos ( \pi x - 2t)= \sin (\pi x - 2t + \frac{\pi}{2})

I might have interpreted your equation wrong, but you get the idea. The phase velocity is still w/k, whether it's sin or cos.

I probably should have used a variable other than 'x' when I wrote that identity. Don't confuse it with the same x in your equation. I just meant that cos of anything is the same as sin plus an extra pi/2 phase shift.

Oh, and you missed a '2' before the trig function :)

It is a pain when there is no edit button, isn't it? =/

Otherwise, yes, in any form, if the function was sine instead of cosine, the speed would be the same. The only thing that changes is the path difference which differs by a quarter of a cycle.

Yeah, I did miss that pesky 2, didn't I? Oops! Thanks.