Motion in a force field

1. Aug 18, 2006

Klarinettus

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.

2. Aug 20, 2006

benorin

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 $$\frac{d^2x}{dt^2}=a(t)$$ is up with the LaTeX?

Last edited: Aug 20, 2006
3. Aug 20, 2006

Klarinettus

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
4. Aug 20, 2006

quasar987

What Newton's second law says is that given the force field $\vec{F}(\vec{r})$ asigning a force to every point in space, the path $\vec{r}(t)$ 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:)

$$\vec{F}(\vec{r(t)}) = m\frac{d^2\vec{r}}{dt^2}(t)$$

5. Aug 20, 2006

qbert

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).