Using Differential Equations to Solve for Particle Motion in a Force Field

[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.

Any advice?

 

[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?

 

[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.

 

[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:)

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

 

[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).