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Homework Help: (x sub 0)(cosine theta)

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data
    At 16:30 he writes (x sub 0)(cosine theta). Is x sub 0 the vertical component length of the position vector of the particle? When he labels x sub 0 a little while before, it looks as though x sub 0 is at the radial point along the x axis. How can you use x sub 0 cosine theta at the radial point along the x axis when the position of the particle doesn't coincide with the point at axis?


    Also, I know this doesn't adhere to the default question set-up guidelines, but how would I go about properly asking a question like this? I don't do schoolwork. I study what my weak mind can grasp alone at a pace at which I find most comfortable. I really don't know what he meant by this.
    Last edited: Sep 11, 2013
  2. jcsd
  3. Sep 11, 2013 #2


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    x = xocos(wt + δ) describes SHM with maximal amplitude xo(since cos never exceeds 1). Since the cos function is periodic, he represented the function as a circle with radius xo, the max amplitude. The projection of xo onto the x axis is the xocosθ.

    This is similar to how sin and cos are defined via the unit circle, except here the circle is not of radius unity.
  4. Sep 11, 2013 #3
    But how is the projection of the maximal amplitude on the the x axis xocosθ? I thought you use cosine when you want to determine vector components that make up the position of a point along the circle.

    He says in the video that to determine the position of the particle along the x axis, you need to take the maximal amplitude times the cosine of theta. Why? What does radius have to do with the position along the x axis which outlines both vector components? Can you try to break this down and explain the intuition even further?
  5. Sep 11, 2013 #4
    Wait does x_0 apply to all radii within the circle? Did he just indicate that the radius, not at the specific spot along the x axis, is x_0? If this is the case, I understand my confusion. And the formula just happens to work now.
  6. Sep 12, 2013 #5


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    Yes, any point lying on the circle is a distance xo away from the origin.
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