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TomRiddle369 · Jan 28, 2011, 08:43 AM

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

Unknown008 · Jan 28, 2011, 09:31 AM

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

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

?

jcaron2 · Jan 28, 2011, 09:39 AM

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.

jcaron2 · Jan 28, 2011, 09:42 AM

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.

Unknown008 · Jan 28, 2011, 09:50 AM

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.

jcaron2 · Jan 28, 2011, 01:22 PM

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