1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Motion in a force field

  1. Aug 18, 2006 #1
    I wasn't sure whether to put this in the math section or the physics section because it's a bit of an overlap problem.

    I want to know how to find the position as a function of time of a particle given its acceleration as a function of position. I know this is some sort of differential equation but I'm confusing myself with it.

    Any advice?
     
  2. jcsd
  3. Aug 20, 2006 #2

    benorin

    User Avatar
    Homework Helper

    If a(t) and x(t) are the acceleration and position of the particle as a function of time, then x''(t)=a(t)

    Edit: What the [tex]\frac{d^2x}{dt^2}=a(t)[/tex] is up with the LaTeX?
     
    Last edited: Aug 20, 2006
  4. Aug 20, 2006 #3
    You're not understanding what I'm asking.

    You are not given a(t). You are given a field assigning a force or acceleration to every point in space.

    Also, you're given the particle's initial position and velocity.

    The goal is to find the position as a function of time.

    To make things simple, let's just look at this situation for motion along a line.
     
    Last edited: Aug 20, 2006
  5. Aug 20, 2006 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What Newton's second law says is that given the force field [itex]\vec{F}(\vec{r})[/itex] asigning a force to every point in space, the path [itex]\vec{r}(t)[/itex] of a particle of mass m in this force field is a solution of the differential equation: (or rather of the 3 following coupled ode:)

    [tex]\vec{F}(\vec{r(t)}) = m\frac{d^2\vec{r}}{dt^2}(t)[/tex]
     
  6. Aug 20, 2006 #5
    As quasar said, you just solve the system
    of differential equations. In general you're going
    to have to do it numerically.

    For simple 1-D potentials, though, you use
    a standard trick

    a = dv/dt = (dv/dx)(dx/dt) = v (dv/dx).

    So for the differential equation
    ma = f(x)
    we have
    m v (dv/dx) = f(x).
    mv^2/2 - mv0^2 /2 = integral(f(s), s=x0..x) .

    Now put g(x) = sqrt( v0^2 + 2/m int(f(s), s=x0..x) )
    so that we gave
    v = g(x)

    then you have v = dx/dt = g(x)
    which is seperable so that

    int( 1/g(s), s = x0.. x) = t - t0

    you invert this to get x = x(t).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Motion in a force field
  1. Force Fields and Curl (Replies: 3)

  2. Work done by Force Field (Replies: 11)

Loading...