1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find a function of velocity relative to x
Loading...