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How to prove that position by velocity is a constant vector

  1. Jun 28, 2015 #1
    [Note from mentor: this thread was originally posted in a non-homework forum, therefore it does not use the homework template.]

    -------------------------------

    So I've been asked to prove that in a harmonic function where

    a(t)+w2r(t)=0

    that

    (1) v(t).v(t)+w2r(t).r(t)=constant scalar

    and

    (2) r(t).v(t)=constant vector

    where a(t)=acceleration, v(t)=velocity, r(t)=position


    By deriving (1) I found that

    2[a(t)+w2r(t)].v(t)=0 because a(t)+w2r(t)=0

    By deriving (2) I get

    v(t).v(t)+r(t)a(t)= v(t).v(t)+r(t)[-w2r(t)] because a(t)=-w2r(t)

    How do I finish this?

    Can anyone please explain what the point of this proof is?

    Thanks!
     
    Last edited by a moderator: Jun 28, 2015
  2. jcsd
  3. Jun 28, 2015 #2
    What is v(t) in terms of the time derivative of r(t)? What is a(t) in terms of the time derivative of v(t)?

    Chet
     
  4. Jun 28, 2015 #3

    Fredrik

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    Gold Member

    r⋅v isn't a vector. Is that perhaps supposed to be a cross product?

    I don't think r⋅v is constant.
     
  5. Jun 28, 2015 #4
    Just solve the ODE, you'll get values of r and a, verify if these hold when plugging them in eq 1 and 2, Cheers!
     
  6. Jun 28, 2015 #5
    Thanks for replying

    It doesn't actually v(t) in terms of r(t) or a(t) in terms of v(t). I'm meant to show that it's true for all simple harmonic oscillators
     
  7. Jun 28, 2015 #6
    I know that. But, you can derive your result for part 1 by using these relationships, multiplying your starting equation by v, and integrating with respect to t. It's really simple.

    Chet
     
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