# Solve for x(t), I just keep seeing 0

1. Feb 28, 2008

1. The problem statement, all variables and given/known data
$$\frac{dx}{dt} = \sqrt{2(\frac{E}{m}) -\omega^2 x^2}$$

2. Relevant equations

$$\frac{dx}{dt} = v$$

3. The attempt at a solution

I keep ending up with the statement x=0.

$$\frac{dx}{dt}=\sqrt{2\frac{E}{m} -\omega^2 x^2}$$

Then
$$(dx/dt)^2 =2\frac{E}{m} -\omega^2 x^2$$

and 1/2 mv^2 = E

$$(dx/dt)^2 = v^2 - 2 \omega^2 x^2$$

and with dx/dt being v that makes

$$0=\omega^2x^2$$

2. Feb 28, 2008

### kdv

What is exactly the question you are trying to answer?

In any case, what is E here? If it's th etotal energy, it's not simply 1/2 mv^2 since there is some potential energy.

3. Feb 28, 2008

Just find x(t)

I left off a 2 in my work that I just noticed and that seems to have changed things a bit. I have to rework some stuff and maybe see where that ends up. As for what E is, I don't know. Nor do I technically know that m is mass, but it seems awfully like they are. They weren't explicitly defined.

Last edited: Feb 28, 2008
4. Feb 28, 2008

### kdv

Ok, then don't assume E is 1/2 mv^2

This is a simple harmonic oscillator with E = 1/2 mv^2 + 1/2 k x^2.

5. Feb 28, 2008

### HallsofIvy

Staff Emeritus
If the problem is given just as the differential equation, you don't know what E is nor do you know that m is mass, how do you possibly conclude that E= (1/2)mv2?
(And why is this a physics problem rather than a math problem?)

6. Feb 28, 2008

The best way to attack th eproblem is to differentiate your equation to get the second derivative $$\frac{d^2x}{dt^2}$$ Use the initial equation to rewrite the first derivative that appears in your result in terms of x itself. Then you will have an equation relating the second derivative to the function x(t) and you will be able to solve.