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DU/dt = 0 for oscillating spring, help with derivation

  1. Dec 6, 2012 #1
    U = energy
    In the book:
    [itex] \frac{dU}{dt} = \frac{d}{dt} (\frac{1}{2} mv^2 + \frac{1}{2} kx^2) [/itex]

    then we have [itex] m \frac{d^{2}x}{dt^2} + kx = 0 [/itex] because [itex] v = \frac{dx}{dt} [/itex]

    however they get rid of [itex] \frac{dx}{dt} [/itex] .

    They are ignoring the case where v = 0, because then [itex] m \frac{d^{2}x}{dt^2} + kx [/itex] doesn't have to be zero, and it can still satisfy the equation.
    Last edited: Dec 6, 2012
  2. jcsd
  3. Dec 6, 2012 #2


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    Homework Helper

    If you have y = (dx/dt)^2 and you put u = dx/dt

    then y=u^2 such that dy/du = 2u and du/dt = d^2x/dt^2

    So dy/dt = 2u*du/dt = 2(dx/dt)(d^2x/dt^2)

    In your original equation, differentiating the KE term and the spring term will give you a dx/dt which can be canceled out since dU/dt= 0.
  4. Dec 6, 2012 #3
    You should edit the post and replace [; ... ;] with [i tex] ... [/i tex]
    (get rid of the space in [i tex]. I put that in so the parser wouldn't detect it.)
  5. Dec 6, 2012 #4
    yes, i get it but if dx/dt = 0, which it can, then the equation is satisfied and the other term doesn't have to be zero. However, we are saying the other term must always be zero.
  6. Dec 7, 2012 #5


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    Staff: Mentor

    dx/dt = 0 is true for a two point in time per period only, or for a non-moving spring in equilibrium. That is not relevant for the general case.
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