# Period of an Oscillating Particle

1. Jun 11, 2013

### robertjordan

1. The problem statement, all variables and given/known data

A particle oscillates with amplitude A in a one-dimensional potential that is symmetric about x=0. Meaning U(x)=U(-x)

First find velocity at displacement x in terms of U(A), U(x), and m.

Then show that the period is given by $4\sqrt{\frac{m}{2U(A)}}\int_0^A \frac{1}{1-\frac{U(x)}{U(A)}}dx$
(hint: $dt=\frac{dx}{v}$)
2. Relevant equations

Total energy = $U(x) + \frac{1}{2}mv^2$
Period = time it takes to complete an oscillation

I found velocity by noting that at x=A, all the energy is potential, so total energy of the system (which doesn't change) is U(A), so we can set up the equation:

$U(A)=U(x)+\frac{1}{2}mv^2$ and solving for v we get:

$v=x'=\sqrt{\frac{2(U(A)-U(x))}{m}}$.

3. The attempt at a solution

Now I was thinking we could multiply the velocity at x=0 by Δt, then do v(0)Δt+v(Δt)Δt, and then v(0)Δt+v(Δt)Δt+v(2Δt)Δt and so on and so forth... this will help us find how far the particle has traveled at any time t. But obviously the right way to do the above process is by integrating our $v$ equation with respect to time, which will give us distance traveled over that time which will let us know how long it takes to travel A distance (1/4 of an oscillation).

$\int_0^t v dt = \int_0^t \sqrt{\frac{2(U(A)-U(x))}{m}}dt$

But we know $\frac{dx}{dt}=v$ so we can write $dt=\frac{dx}{v}$...

umm... I'm stuck...

Last edited: Jun 11, 2013
2. Jun 11, 2013

### voko

You have a differential equation in the form $\frac {dx} {dt} = f(x)$. How do you integrate it?

If unclear, try these simple examples: $f(x) = a + x$, $f(x) = x^2$.

3. Jun 11, 2013

### robertjordan

Since the input into our velocity equation we found is x not t, I should edit:

... to say:

See how as we make Δt smaller and smaller, the above becomes a better and better approximation of displacement x as a function of time? But I don't know how to turn that method into an integral?

As for integrating $\frac{dx}{dt}=f(x)$, if we do $\int \frac{dx}{dt}dt=\int f(x)dt$, we get $x(t)=\int f(x)dt$... (don't know how we could evaluate that integral.)

Whereas if we do it $\int \frac{dx}{dt}dx = \int f(x)dx$ I don't know what the left side would be and we'd have to evaluate the right side.

How are either of these useful to me?

4. Jun 11, 2013

### voko

Have you studied differential equations? Separation of variables sound familiar?

5. Jun 11, 2013

### robertjordan

$\frac{dx}{dt}=f(x)$
$\frac{1}{f(x)}dx=dt$
$\int f(x)^{-1} dx = \int 1 dt=t$

is that right?

That would give us time as a function of velocity which can give us velocity as a function of time?

Then we can integrate that with respect to t to find displacement as a function of time?

thanks for the quick response btw

Last edited: Jun 11, 2013
6. Jun 11, 2013

### voko

What would that have to do with velocity? Is x velocity?

7. Jun 11, 2013

### robertjordan

oh whoops the integral of velocity is displacement so we would have x(t). So we could solve x(?)=A and then multiply it by 4 to get the period?

8. Jun 11, 2013

### voko

You will have t(x) for sure. Whether you can convert that to x(t) depends on U(x).

But if you just need to get the period, you don't need x(t). t(x) is exactly what you need in this case.

9. Jun 11, 2013