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Homework Help: Solve for x(t), I just keep seeing 0

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\frac{dx}{dt} = \sqrt{2(\frac{E}{m}) -\omega^2 x^2}[/tex]

    2. Relevant equations

    [tex] \frac{dx}{dt} = v [/tex]

    3. The attempt at a solution

    I keep ending up with the statement x=0.

    [tex]\frac{dx}{dt}=\sqrt{2\frac{E}{m} -\omega^2 x^2}[/tex]

    [tex](dx/dt)^2 =2\frac{E}{m} -\omega^2 x^2 [/tex]

    and 1/2 mv^2 = E

    [tex](dx/dt)^2 = v^2 - 2 \omega^2 x^2[/tex]

    and with dx/dt being v that makes


    and then I'm sad :-(
  2. jcsd
  3. Feb 28, 2008 #2


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    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.
  4. Feb 28, 2008 #3
    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
  5. Feb 28, 2008 #4


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    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.
  6. Feb 28, 2008 #5


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    Science Advisor

    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?)
  7. Feb 28, 2008 #6
    At the time I felt that it was more of a physics problem.
  8. Feb 28, 2008 #7


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    The best way to attack th eproblem is to differentiate your equation to get the second derivative [tex] \frac{d^2x}{dt^2} [/tex] 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.
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