Finding an expression for x(t)

[crimpedupcan]

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)?

 

[mikeph]

integrate!

v(t) = integral of a(t)dt + v(t=0)
x(t) = integral of v(t)dt + x(t=0)

 

[crimpedupcan]

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)?

 

[mikeph]

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.

 

[crimpedupcan]

Thanks a lot!
I think I've seen that method applied before to simple harmonic motion, I wish I made the connection earlier.

 

[mikeph]

No problem. It caught me a little off guard as well!

 

[phyzguy]

If you only know a(x), you can use the following:
$$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx} = a(x)$$
Then:
$$\int v dv = \frac{v^2}{2}= \int a(x) dx$$
After integrating, this can be solved for v(x). Then:
$$\frac{dx}{dt} = v(x)$$
$$\int dt = t = \int \frac{dx}{v(x)}$$
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.

 

[crimpedupcan]

Thank you, that's exactly what I was looking for

 

[stevendaryl]

[deleted]

I just repeated the same thing that phyzguy said.