Wave direction for Asin(kx±wt) and Acos(kx±wt)

[EV33]

Homework Statement

I don't have an actual problem but is what I am trying to make sure I have clear is...

What direction in general is
Asin(kx+or-wt) going?


Asin(kx+wt) Left
Asin(kx-wt) Right
Asin(-kx-wt)=-Asin(kx+wt) Left
Asin(-kx+wt)=-Asin(kx-wt) Right

Acos(kx+wt) Left
Acos(kx-wt) Right
Acos(-kx+wt)?
Acos(-kx-wt)?

Homework Equations

Asin(kx-wt) is to the right
Asin(kx+wt) is to the left

Is what I have been taught to help me figure out the direction.

The Attempt at a Solution

I am pretty sure the ones I answered are correct but if they're please let me know.

My trouble is that as far as I know I can't take a negative out of the cos function like I can the sin function, so the only thing I have thought of thus far...

Acos(-kx+wt)=Asin(-kx+wt+pi/2)=-Asin(kx-wt-pi/2)

So I would say this goes to the right because it is minus wt.

Does that work?

 

[ehild]

Cosine is an even function, cos(-x)=cos(x) so

Acos(-kx+wt)=Acos(kx-wt)
Acos(-kx-wt)=Acos(kx+wt).

You can visualize the traveling of a wave by picking up a crest and seeing in what direction and with what speed it is moving.

If it is cosine A(wt-kx), a crest is at wt-kx=0. This means that at t= 0 the crest is at x=0 and at a later time t, it is at x= k/w*t. This is the same equation that holds for a body moving along the x-axis with uniform velocity v=k/w in the positive direction (from left to right). In case of Acos(wt+kx), the crest is at x=-w/k*t at time t, so the crest moves in the negative direction, from right to left.

For a sine wave, Asin(wt-kx), a crest appears where wt-kx =pi/2. At t = 0 it is at x=-pi/(2k) and it moves according to the equation x= w/k*t-pi/2 that is, towards higher x values, from left to right. In case of Asin(wt+kx), the crest moves according to wt+kx=pi/2, that is x=pi/(2k)-w/k*t, in the negative direction.

ehild