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F(t) as it relates to v(t) and x(t)

  1. Oct 11, 2006 #1
    Starting at t = 0, an object of mass m is subject to a force

    F(t) = F nought cos omega t

    If the initial speed and position are v nought and x nought, find v(t) and x(t), any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 11, 2006 #2

    quasar987

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    Well, now you gotta solve

    [tex]\frac{dv}{dt}=\frac{F_0}{m}\cos(\omega t)[/tex]

    That's easy. What is the function whose derivative is cos?
     
  4. Oct 11, 2006 #3
    Okay, so this is how it worked out for me:

    dv = (F0/m) cos(ωt) dt

    v = (F0/mω) sin(ωt) + C
    v(t) = (F0/mω) sin(ωt) + v0
    dx =[ (F0/mω) sin(ωt) + v0 ] dt
    x = -(F0/mω2) cos(ωt) + v0t + x0

    x(t) = -(F0/mω2) cos(ωt) + v0t + x0

    Does that make sense?
     
  5. Oct 11, 2006 #4

    quasar987

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    You went a little too fast. After integrating

    dx =[ (F0/mω) sin(ωt) + v0 ] dt

    you get

    x = -(F0/mω²) cos(ωt) + v0t + C

    But plugging x(0)= x_0 does not give C=x_0 because cos(0)=1.
     
  6. Oct 11, 2006 #5
    Oh, oops, so then C = x_0 + (F_0/mω²) ?
     
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