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Choosing what's integrated

  1. Jan 9, 2007 #1
    Sorry for the undescriptive title, but I couldn't think of a better one.

    My question is essentially this: is the following procedure correct?

    a=-kx and we want v.

    Multiply both sides by v to get:

    [tex] \frac {d^2x} {dt^2} * \frac {dx} {dt} = -kx *\frac {dx} {dt} [/tex]

    Now, by dt, and some rearranging :

    [tex] (\frac {d^2x} {dt^2} dt) * \frac {dx} {dt} = -kx *dx [/tex]

    And the step which I'm not convinced can be done:

    [tex]( \int \frac {d^2x} {dt^2} dt) * \frac {dx} {dt} = -k \int x *dx [/tex]

    and so we end up with v^2 = -(kx^2)/2 +c which is but an errant 1/2 away from what I need.

    I just don't trust integrating a wrt t, while having a v just sitting there as if it were a constant. The only reason that I haven't already disregarded it is that it checks out if you analyse the dimensions...
  2. jcsd
  3. Jan 9, 2007 #2


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    Homework Helper

    No, that's not right. As you mention, you're treating v as a constant, and it isn't.

    You started out right, you just need to rewrite:

    [tex] \frac {d^2x} {dt^2} * \frac {dx} {dt} [/tex]


    [tex]\frac{1}{2} \frac{d}{dt} \left( \left(\frac{dx}{dt} \right)^2 \right)[/tex],

    then integrate.
  4. Jan 9, 2007 #3
    Wow, quite the nifty identity. Thanks a lot for that help.
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