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Where is this equation derived from? (general physics)

  1. Dec 12, 2011 #1
    Hey, I was wondering what equations the following equation is derived from:

    v=w√(A2-x2)

    I figure v=wA is one of them but I'm not sure how A2=A-x

    Thanks.

    edit: The only reason I don't think this is an equation I have to memorize, but one that I can build from equations I already have memorized.
     
    Last edited: Dec 12, 2011
  2. jcsd
  3. Dec 12, 2011 #2

    e.bar.goum

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    I think you'll have some more luck on the forum if you defined your terms. v, w, A and x can mean many different things in physics.
     
  4. Dec 12, 2011 #3
    velocity, angular velocity, amplitude, distance
     
  5. Dec 12, 2011 #4

    Delphi51

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    Well, in rotational motion, v =r*ω. Perhaps it is for a situation where that square root gives you the radius.

    Oh, you say A is amplitude. So it must be something to do with a wave or periodic motion.

    In physics we are mostly in the business of studying a situation, then finding the formula that describes it.
     
  6. Dec 12, 2011 #5

    cepheid

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    Assuming that you're talking about simple harmonic motion in the x-direction with angular frequency (not angular velocity) ω such that the x position vs. time is given by the equation

    x = Acos(ωt)

    then the velocity in the x-direction vs. time is given by (and you just have to accept this for now):

    v = -ωAsin(ωt)

    Now, ask yourself, what is x2 + v2 equal to? Can you see how to simplify this using a special property of sine and cosine? Hint: it helps if you modify the v term by dividing out a certain factor first. The end result should be the equation that you're trying to derive.
     
  7. Dec 12, 2011 #6

    cepheid

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    I screwed up the equation for the velocity in my previous post and have since edited it.
     
  8. Dec 13, 2011 #7

    cepheid

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    I also just realized that, for SHM of a horizontal spring, in particular, there is a much easier way of deriving this using the conservation of energy. It doesn't require you to know the explicit time-dependence of the variables (i.e. you don't need to know the expressions for x and v vs. time using this method).
     
  9. Dec 13, 2011 #8
    so using conservation of energy .5kA^2=.5mv^2

    v=√(A^2k/m)
    w^2=k/m

    v=√(w^2A^2)

    That yields v=wA...

    I'd appreciate a little more help.

    Nevermind, using .5mv^2+.5kx^2=.5kA^2 works.

    Thanks.
     
  10. Dec 13, 2011 #9
    If you watch a bouncing spring (that has the same behavior when stretching and compressing) you will notice that it stops where x = A and is going the maximum speed where x = zero and your original formula is consistent with this, so this may help you remember it.. Turn circular motion on its edge and you have the bouncing spring situation so your angular velocity relation to linear veloctiy holds as do the energy equations.
     
  11. Dec 13, 2011 #10
    If you want to see where the trig expression in SHM come from you should consider the circular motion that can be associated with every SHM.
    The attached sketch shows a point P moving in circular motion with constant
    angular velocity ω.
    The point N is the projection of P onto a diameter and N executes SHM. In the sketch the SHM displacement is x.
    All of the SHM equations can be derived from the circular motion equations and, more importantly, can be 'seen'
    I hope you can see that the SHM velocity is the 'horizontal' component of the circular motion velocity = VSin∅ .... Sin∅ = d/r (d has no meaning for the SHM).
    I hope that this geometry helps to see where SHM equations come from !!!!
    SHM is circular motion viewed edge on.... just as netgypsy observes.
     

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    Last edited: Dec 13, 2011
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