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Rearranging Derivatives

  1. Sep 4, 2004 #1
    I've got this from a book. My question is simple. Does anyone know how to get to the expression (UNCLEAR) from Eq. (24.2)? I can verify it without any trouble, but the author does not explain how it was obtained in the first place.

    "Now let us consider the energy in a forced oscillator. The equation for the forced oscillator is
    [tex]m\frac{d^2 x}{dt^2}+\gamma m \frac{dx}{dt}+m\omega _0 ^2 x = F(t). [/tex] (24.1)

    In our problem, of course, [tex]F(t)[/tex] is a cosine function of [tex]t[/tex]. Now let us analyze the situation: how much work is done by the outside force [tex]F[/tex]? The work done by the force per second, i.e., the power, is the force times the velocity. [tex]\Big([/tex] We know that the differential work in a time [tex]t[/tex] is [tex]F dx[/tex], and the power is [tex]F\frac{dx}{dt}[/tex]. [tex]\Big)[/tex] Thus

    [tex]P=F\frac{dx}{dt}=m\left[\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)+\omega _0 ^2 x\left(\frac{dx}{dt}\right) \right]+\gamma m\left(\frac{dx}{dt}\right)^2 .[/tex] (24.2)

    But the first two terms on the right can also be written as
    [tex]\frac{d}{dt}\left[\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 +\frac{1}{2}m\omega _0 ^2 x^2 \right], [/tex] (UNCLEAR)
    as is immediately verifyed by differentiating."
  2. jcsd
  3. Sep 4, 2004 #2


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    Note that
    [tex]\frac{d^2 x}{dt^2} = \frac {d}{dt} \frac {dx}{dt}[/tex]
    and, generally,
    [tex]g \frac {dg}{dt} = \frac {1}{2} \frac {d g^2}{dt}[/tex]
    where g = g(t) is any function of t.
    Last edited: Sep 4, 2004
  4. Sep 5, 2004 #3
    That works!! Thank you.

    Thank you for your reply, Tip. That [tex]g \frac {dg}{dt} = \frac {1}{2} \frac {d g^2}{dt}[/tex] pattern seems to fit!
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