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X = Acos(ωt) + Bsin(ωt) derivation

  1. Mar 11, 2012 #1
    How do you derive x = Acos(ωt) + Bsin(ωt) from F = -mω2x and what is the former used for?

    Thank you!!
     
  2. jcsd
  3. Mar 11, 2012 #2

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    Hey sparkle!

    If this is homework, you should show some effort before we're allowed to help you (PF regulations I'm afraid).
    What's it for?
     
  4. Mar 11, 2012 #3
    This was on a list of things you should know for physics contests. :)
     
  5. Mar 11, 2012 #4

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    Well, what can you make of it?
     
  6. Mar 11, 2012 #5
    Well, F = -mω2x is hooke's law
    and SHM for a spring is like x = Asin(ωt)
     
  7. Mar 11, 2012 #6

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    So what's your question?

    Actually, F=ma and "a" is the second derivative of "x" with respect to time.
    So you have mx''=-mω2x.
    The general solution to this differential equation is x = Acos(ωt) + Bsin(ωt).
     
  8. Mar 11, 2012 #7
    so would you get from Acos(ωt) + Bsin(ωt) to Asin(ωt)?
    thanks!
     
  9. Mar 12, 2012 #8

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    The relation is:
    $$A\cos(ωt) + B\sin(ωt) = \sqrt{A^2+B^2}\sin(ωt+θ_0)$$
     
  10. Mar 12, 2012 #9
    thank you! :)
     
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