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Homework Help: Energy Conservation for a spring

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a mass connected to a spring of sti ness k.
    (a) Use conservation of energy to write down a first order differential equation obeyed
    by the mass.
    (b) Find the time t for the mass to move from the origin at t = 0 out to a position
    x assuming that at t = 0 it has initial speed dx/dt = v0.
    (c) Use the above to show that the period of motion is T = 2∏sqrt(m/k)

    2. Relevant equations

    Us = .5kx^2
    K = .5mv^2

    3. The attempt at a solution

    a) I just set up a initial and final conservation of energy

    I figured that I'd just set Us = K and make v = (dx/dt). That'd give me a natural log in my answer, which doesn't seem to make much sense.

    Everything else doesn't seem to be too hard once I get part a.
  2. jcsd
  3. Feb 28, 2013 #2

    Simon Bridge

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    I'd just set Us = K and make v = (dx/dt)

    That would be: $$\frac{dx}{dt} = \pm \sqrt{\frac{k}{m}}x$$ ... if you are looking for x(t) then the solution should be an exponential, not a log.

    Notice that you can have a positive or a negative velocity for any given position.
  4. Feb 28, 2013 #3
    Wouldn't that set up yield this...

    dx/x = ±sqrt(k/m)*dt

    and if you integrate...

    ln(x) = ±sqrt(k/m)*t

    I don't know much about eulers identity, but what if I took an exponential of that answer. Then it'd be

    x = e^[±sqrt(k/m)*t]

    which (and I know this part is wrong) I'd write as

    x = cos[sqrt(k/m)*t]

    If it were second order, then I could get the trig function that I think is supposed to happen from this with an ODE.


    I feel like the problem lies in my first equation. Any thoughts on that?
    Last edited: Feb 28, 2013
  5. Feb 28, 2013 #4


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    Yes, the problem lies in the first equation. Energy is conserved means Us+K=C where C is some constant. Write that out then take the time derivative of both sides. That will give you a second order ode.
    Last edited: Feb 28, 2013
  6. Feb 28, 2013 #5

    Simon Bridge

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    What Dick said - the clue is that you have not accounted for the ##\pm## ... didn't work: I should have just directed you to look again at what "conservation law" means ;)
  7. Mar 1, 2013 #6
    The only thing I'm worried about is that the question specifically says to write down a first order ODE, but I guess it doesn't make much sense in this case.

    So how do I take a time derivative of that? I'm a little confused because of the (dx/dt)^2
    I'm guessing I want kx + ma = 0

    wouldn't the chain rule get

    d/dt(.5kx^2 + .5m(dx/dt)^2)) = kvx + mva = 0
    Last edited: Mar 1, 2013
  8. Mar 1, 2013 #7

    Simon Bridge

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    ... well done: cancel common factor?
    Note: you have written down a 1st order ODE - this is the next step.

    You could always have done: $$\frac{dx}{dt} = \pm \sqrt{c-\frac{kx^2}{m}}$$
    ... but you'd still have to figure how the ##\pm## should be handled.
    Last edited: Mar 1, 2013
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