How to prove that position by velocity is a constant vector

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heenac2
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[Note from mentor: this thread was originally posted in a non-homework forum, therefore it does not use the homework template.]

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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!
 
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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!
 
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Chestermiller said:
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

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
 
heenac2 said:
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
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