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Finding an expression for x(t)

  1. Mar 19, 2013 #1
    Suppose I have a particle on a line, and I know some function a(x) and the initial x, v, and a. How could I work out x(t)?
  2. jcsd
  3. Mar 19, 2013 #2

    v(t) = integral of a(t)dt + v(t=0)
    x(t) = integral of v(t)dt + x(t=0)
  4. Mar 19, 2013 #3
    Thanks for the reply.
    From what I understand your solution requires me to know a(t), but what can I do if I only know a(x)?
  5. Mar 19, 2013 #4
    Ah, then you have:

    x''(t) = f(x(t))

    This is a 2nd order autonomous differential equation with a general solution, although if f is tricky then you might need a numerical integrator to solve it. The trick is to multiply through by 2x'(t) then factor the left side into ((x'(t))^2)'. Then you can find x(t) by integrating, taking the square root, and integrating again.
  6. Mar 19, 2013 #5
    Thanks a lot!
    I think I've seen that method applied before to simple harmonic motion, I wish I made the connection earlier.
  7. Mar 19, 2013 #6
    No problem. It caught me a little off guard as well!
  8. Mar 19, 2013 #7


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    If you only know a(x), you can use the following:
    [tex]\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx} = a(x)[/tex]
    [tex]\int v dv = \frac{v^2}{2}= \int a(x) dx[/tex]
    After integrating, this can be solved for v(x). Then:
    [tex]\frac{dx}{dt} = v(x)[/tex]
    [tex]\int dt = t = \int \frac{dx}{v(x)}[/tex]
    After integrating, this gives t(x), which can then be inverted to give x(t). This procedure can be complicated and mathematically difficult, but it will work, at least numerically.
  9. Mar 19, 2013 #8
    Thank you, that's exactly what I was looking for
  10. Mar 19, 2013 #9


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    I just repeated the same thing that phyzguy said.
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