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Homework Help: Find a function of velocity relative to x

  1. Nov 2, 2013 #1
    1. The problem statement, all variables and given/known data

    An object is moving in x-axis with given force [itex]Fx = -mω^{2}x[/itex], where ω is a positive constant. when t = 0 the object starts moving from point [itex]x_{0} > 0[/itex].
    a. Find the object's velocity function relative to x ([itex]v(x)[/itex])
    b. Find the object's position function relative to t ([itex]x(t)[/itex])

    Help: [itex]\int\frac{dz}{\sqrt{A^{2} - z^{2}}}[/itex] perform the integration by changing z = A cos u
    (I don't know what this help thing for though, I find it irrelevant to the question :confused: )

    2. Relevant equations

    3. The attempt at a solution

    a. First, using the above-mentioned force equation, I found the acceleration function relative to x, it goes like:

    [itex]Fx = -mω^{2}x[/itex]
    [itex]ma(x) = -mω^{2}x[/itex]
    [itex]a(x) = -ω^{2}x[/itex]​

    then, since [itex]a(x) = \frac{v(x)}{dt}[/itex], by using chain rule, it goes:

    [itex]a(x) = \frac{v(x)}{dx}\frac{dx}{dt}[/itex]
    [itex]a(x) dx = v(t) dv(x)[/itex]​

    assuming that v(t) = v(x) (even though the function might be different, the result will be the same either way) then:

    [itex]a(x) dx = v(x) dv(x)[/itex]
    [itex]\int a(x) dx = \int v(x) dv(x)[/itex]
    [itex]-\frac{1}{2}ω^{2}x^{2} + C = \frac{v(x)^{2}}{2} + C[/itex]
    [itex]v(x) = \sqrt{-1}ωx[/itex]​

    there goes my attempt for part a, and I'm not even sure if it was correct, that [itex]\sqrt{-1}[/itex] part is throwing me off :uhh:

    for part b, my last-moment attempt goes like this:

    [itex]v(t) = \int a(x) dt[/itex]
    [itex]v(t) = -ω^{2}xt + C[/itex]​

    when t = 0, the object is at rest, then v(0) = 0, and when t = 0, [itex]x = x_{0}[/itex]

    [itex]0 = -ω^{2}x_{0}(0) + C ==> C = 0[/itex]
    [itex]v(t) = -ω^{2}xt[/itex]​

    then using integral to find x(t):

    [itex]x(t) = \int v(t) dt[/itex]
    [itex]x(t) = \frac{-ω^{2}x_{0}t^{2}}{2} + C[/itex]​

    when t = 0,

    [itex]x_{0} = \frac{-ω^{2}x_{0}(0)^{2}}{2} + C ==> C = 0[/itex]
    [itex]x(t) = \frac{-ω^{2}xt^{2}}{2} + x_{0}[/itex]​


    pretty sure I was wrong for part b :s
    can someone give me a hint on how to solve this problem?
    any help is much appreciated, thanks! :smile:
  2. jcsd
  3. Nov 2, 2013 #2


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    Welcome to PF!

    What exactly are you doing? I'm afraid this is nonsense. The acceleration is the time derivative of the velocity. So, it is:$$a(x) = \frac{dv(x)}{dt}$$and so by the chain rule:$$a(x) = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$$
  4. Nov 2, 2013 #3


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    Now thing to do is to look at the right hand side of that last equation, and notice that it looks like something that *has already been* differentiated with respect to x using the chain rule. So, can you reverse that step, thus turn the right hand side into an exact derivative of something? I.e., right hand side can be expressed as ##\frac{d}{dx}[\textrm{blah}]##, and you have to figure out what "blah" is.
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