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About velocity equation in simple harmonic motion

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Actually this is not a problem. This is about something i couldnt understand.

    Y5y45A.jpg

    2. Relevant equations
    dairesel3.jpg
    where wt equals θ

    3. The attempt at a solution
    I can understand that x=r.cosθ it is obvious. But i cant understand why Vx=-V.sin(w.t) When i take a triangle from that circle i end up finding Vx equal to V.sin(w.t) . Where do we get that minus? Also how does he find the angle as 90+wt in that equation? I think it should be 90 - wt

    Note: I dont want the explanation with derivative.
     
  2. jcsd
  3. Oct 27, 2014 #2

    haruspex

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    It's all there in the diagram. The angle from the +ve X axis anticlockwise around to the velocity vector is shown as 90+theta. If that's not clear, project the blue radius line through the particle to cut the angle into two parts.
     
  4. Oct 28, 2014 #3
    Ummm.. I think i couldnt express my problem. I can understand why is that angle is 90 + theta. My problem is different.

    If you take triangle which has a pink vector leftward and green vector which making an angle of alpha with the pink vector (where alpha is 90 - theta) and put the Vy component in to the emptry section of that triangle you will see that Vx equals V.cos(alpha) which is 90 - theta so equation becomes Vx=V.cos(90-theta)

    As i know, that 90 degree in the cosinus will turn it into a sinus but wont add a minus to it since 90 - theta is in the first section of coordinate system.

    Summary: My problem is with that minus... Btw thank you for helping.
     
  5. Oct 28, 2014 #4

    haruspex

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    But in that diagram you have drawn Vx as a left-pointing vector. Since right is positive, you should take Vx as right-pointing. You then see that it is Vcos(90+theta), hence the minus sign.
     
  6. Oct 28, 2014 #5
    Got it. Thanks a lot.
     
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