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Particle motion diff eq

  1. Sep 8, 2005 #1
    I'm given F=kvx. I need to find x(t), k is a positive constant, and the particle passes thru the origin with speed Vo at time t=0.

    I start by rewriting the formula as such:

    dv/dt=k*dx/dt*x

    I am confused now though because I now have dv,dt, and dx. I assume I need to get everything in terms of dx and dt. Can I use seperation of variables or do I need another technique?
     
  2. jcsd
  3. Sep 8, 2005 #2

    quasar987

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    Is v the speed?

    If so, you got F=kvx, which can be rewriten as

    [tex]F = k\frac{dx}{dt}x[/tex]

    and F = ma, which can be rewritten as

    [tex]F = m\frac{d^2x}{dt^2}[/tex]

    Combining the two leads to

    [tex]k\frac{dx}{dt}x = m\frac{d^2x}{dt^2}[/tex]
     
  4. Sep 8, 2005 #3
    Now is this where you have trouble? [itex]kx=mx^{2}[/itex] has solutions x=0 and x=k/m. What does this tell you about the differential equation?
     
    Last edited: Sep 8, 2005
  5. Sep 8, 2005 #4
    Ok, after some tinkering I was able to get:
    [tex]k\frac{dx}{dt}x = m\frac{d^2x}{dt^2}[/tex]
    Since it doesn't appear to be able to be put into a 1st order equation I assume I am out of luck with seperation, or am I?

    Sorry, but I'm not sure what x=0, x=k/m tells me about the equation. Are they linear dependent? The lighter the object the further it travels?

    It's been over 2 years since diff eq and my head is a mess right now. I feel helpless. My teacher keeps saying this is easy, but when you've been away from it for a while, it is very hard. It's not like riding a bike that's for sure.
     
  6. Sep 9, 2005 #5

    HallsofIvy

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    Try using the fact (chain rule) that
    [tex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}[/tex]
    to eliminate t from the equation.

    [tex]m\frac{dv}{dt}= mv\frac{dv}{dx}= F= kxv[/tex]
     
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