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Power Dissipation

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    The instantaneous power dissipated by the damping force in a driven oscillator is [tex]P(t) = f_x v_x = -bv_x ^2[/tex].
    Show that the average power dissipated during one cycle of steady-state motion is [tex]\overline{P} = -\frac{1}{2} b\omega^2 A^2[/tex], where [tex]\omega[/tex] is the driving frequency and [tex]A = |\underline{A}|[/tex] is the oscillation amplitude.

    2. Relevant equations

    3. The attempt at a solution
    I'm attempting to just solve an integral for the average power:

    [tex] \omega/2\pi*\int^{2\pi/\omega}_{0} -bv_x^2 dt[/tex]

    But what is [tex]v_x[/tex]? If [tex]x(t) = \underline{A} e^{i \omega t}[/tex], then [tex]v(t) = i \omega \underline{A} e^{i \omega t} = i\omega x(t)[/tex]. So then I think that [tex]v_x = i\omega[/tex] but this doesn't give me the correct answer when put into the integral. Thanks for the help!
    Last edited: Sep 9, 2009
  2. jcsd
  3. Sep 9, 2009 #2


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    Homework Helper

    Use the real solution,

    [tex] x(t) = \underline{A} cos(\omega t) [/tex] or [tex] \underline{A} sin(\omega t)[/tex].

    The world is real, the displacement or velocity of a vibrating body is a real quantity. The complex formalism is just a tool to make solutions easier. It works for linear relations only.

  4. Sep 9, 2009 #3
    So then if I just use [tex] x(t) = A cos(\omega t) [/tex], then it follows that [tex] v(t) = -\omega A sin(\omega t) [/tex]. Is [tex]v_x[/tex] just dv/dt / dx/dt then? That would make it [tex]\omega cot(\omega t)[/tex] which also does not work in the integral
    Last edited: Sep 9, 2009
  5. Sep 9, 2009 #4


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    Homework Helper

    I do not understand what you are doing. [tex]v_x[/tex] is the same as your v(t). The subscribe "x" means the x component of the velocity, and it is the time derivative of the x component of the displacement, x(t). That is, [tex]v_x = dx/dt[/tex] .

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