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Homework Help: Direction of Wave

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant 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.

    3. 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...


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

    Does that work?
  2. jcsd
  3. Apr 13, 2010 #2


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    Homework Helper

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


    You can visualize the travelling 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.

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