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Calculating A in SHM from x, v & a

  1. Apr 19, 2008 #1
    how can you work out the amplitude knowing only displacement, velocity and acceleration?
    I have been able to work out angular velocity, frequency and period, but I can't work out amplitude without knowing either the spring constant, the mass or the phase angle.
     
  2. jcsd
  3. Apr 19, 2008 #2

    Hootenanny

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    Welcome to PF,

    Why don't you post what you have thus far?

    HINT: What is the velocity of the particle when it is at it's amplitude?
     
  4. Apr 19, 2008 #3
    Thank you, it seems like a nice place :)

    I have these equations

    x=Acos([tex]\omega[/tex]t+[tex]\phi[/tex])
    v=-A[tex]\omega[/tex]sin([tex]\omega[/tex]t + [tex]\phi[/tex])
    a=-[tex]\omega[/tex][tex]^{2}[/tex]cos([tex]\omega[/tex]t+[tex]\phi[/tex]) =[tex]\omega[/tex]x

    I could work out [tex]\omega[/tex] by re arranging the equation for a.

    It's 0, but I'm still left with two unknown variables, [tex]\phi[/tex] and A, am I not?

    edit: why does omega keep showing up as superscript?
     
  5. Apr 19, 2008 #4

    Hootenanny

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    Have you any initial conditions, such as whether it starts from x=A or x=0?
     
  6. Apr 19, 2008 #5
    Sorry, I should have typed up the question from the start:
    I struggled with this for hours and concluded that the question must be wrong, but I went to the homepage of the textbook and downloaded the errata and didn't find anything.
     
  7. Apr 19, 2008 #6

    Hootenanny

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    It seems to me that you have a system of simultaneous equations.

    Which values of [itex]\left(\omega t + \phi\right)[/itex] correspond to the particle being at x=A?
     
    Last edited: Apr 19, 2008
  8. Apr 20, 2008 #7
    when x=A, [itex]\left(\omega t + \phi\right)[/itex]=0

    I still can't solve it though, I just end up with division of 0, can you confirm that it really is solvable?
     
  9. Apr 20, 2008 #8

    alphysicist

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    Hi mudkip9001,

    Try conservation of energy. Equate the energy at the given point to the energy at the amplitude. What do you get?
     
  10. Apr 20, 2008 #9
    but I don't know neither k nor m
     
  11. Apr 20, 2008 #10

    alphysicist

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    You'll be able to get rid of them by using what you know about the acceleration.
     
  12. Apr 20, 2008 #11

    alphysicist

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    What I mean is that although you don't know either k or m, you can find the value of (k/m).
     
  13. Apr 20, 2008 #12
    I just wrote all this, and then figured it out myself just as I finished. But I guess I might as well post t anyway. Thanks for all your help, I'm very relieved now (but I also feel a bit dumb, since I was so sure there was something wrong with the question:grumpy:).

    where I went wrong was that the question is how much further it would go, not the amplitude.
     
  14. Dec 3, 2009 #13
    What you were missing is squaring the quantity m/k (i.e., A = \sqrt((x^2+v^2)/(x/a)^2)
    (Sorry I couldn't write this in pretty-print. I'm new here.)
     
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