Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Feynman Lectures, Chapter 24: Transients

  1. Sep 3, 2004 #1
    Hello everyone!

    A question came up as I was reading Chapter 24 of the Feynman Lectures book. To more specific, it's the comments after Eq. (24.2) on the first section---called "the energy of an oscillator". I don't quite get it.

    Thank you very much! :smile:

    "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 analyse 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]
    as is immediately verifyed by differentiating."
  2. jcsd
  3. Sep 3, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    OK, let u = dx/dt. Considering that 1/2 and m are constants, applying the chain rule we can write.

    [tex]\frac{d}{dt}\left[\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 \right] = \frac{1}{2}m\frac{d}{dt}u^2 = \frac{1}{2}m 2 u\frac{du}{dt} [/tex]

    but [tex] \frac{du}{dt} = \frac{d^2x}{dt^2} [/tex]

    so we wind up with

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

    The other half of the problem is similar, substitute as needed for clarity and then apply the chain rule.
  4. Sep 3, 2004 #3
    I see... Thanks!

    Great... I now understand it.

    Thank you so much!
  5. Sep 3, 2004 #4
    Oh... one more thing

    Now I see how the last expression can be expanded. But, How did he get it out of Eq. (24.2)?

    Thank you.
  6. Sep 4, 2004 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)[/tex] is fairly standard.

    If we write u= [tex]\frac{dx}{dt}[/tex], this is just uu'.. but that's just [tex]\frac{1}{2}u^2'[/tex].

    In other words, [tex]\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)= \frac{1}{2}\frac{d\left(\frac{dx}{dt}\right)^2}{dt}[/tex].
    Last edited: Sep 4, 2004
  7. Sep 4, 2004 #6
    Thanks for your tip, HallsofIvy...

    I can understand that it comes down to writing [tex]u\cdot u' .[/tex] The only consequence I can draw from it is that [tex]\frac{dx}{dt}\frac{d\left(\frac{dx}{dt}\right)}{dt}[/tex]. My point is... How does the [tex]\frac{1}{2}[/tex] factor and [tex]\left(\frac{dx}{dt}\right)^2[/tex] appear in [tex]\frac{1}{2}\frac{d\left(\frac{dx}{dt}\right)^2}{dt}[/tex] if we only have [tex]\frac{dx}{dt}[/tex] and [tex]\frac{d^2x}{dt^2}[/tex] to begin with? How do you find it if you initially have [tex]\frac{dx}{dt}\frac{d\left(\frac{dx}{dt}\right)}{dt}[/tex]?
  8. Sep 4, 2004 #7
    Let [itex]u = \frac{dx}{dt}[/itex]. Now
    [tex]\frac{d(\frac{1}{2} u^2)}{du} = u [/tex]
    [tex]\frac{1}{2} \frac{d({\frac{dx}{dt})^2}}{dt} = \frac{du}{dt} \frac{d({\frac{1}{2} u^2)}{du} = \frac{du}{dt} u = \frac{\frac{dx}{dt}}{dt} \frac{dx}{dt} = \frac{d^2{x}}{{dt}^2} \frac{dx}{dt} [/tex]
    Last edited: Sep 4, 2004
  9. Sep 5, 2004 #8
    I finally get it!

    The pattern [tex]g \frac {dg}{dt} = \frac {1}{2} \frac {d g^2}{dt}[/tex] fits for both terms. I finally worked it out. Thanks to HallsofIvy, speeding electron, and Tide for all the help!!!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?