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Driven Harmonic Oscillator: Proving that the max power is given by ω_r = ω_0

  1. Jan 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that that the power given by [itex] \bar{P} = \frac{1}{2} \gamma m \omega_r^2 A_{(\omega)}^2 [/itex] is at a maximum for [itex] \omega_r = \omega_0 [/itex]

    Only variable is [itex] \omega_r [/itex]

    [itex] \omega_r [/itex] is the resonant frequency of the external force while [itex] \omega_0 [/itex] is the eigen frequency of the system.


    2. Relevant equations

    [tex] A_{(\omega)} = \frac{\frac{F_0}{m}}{\sqrt{{(\omega_0^2-\omega_r^2)}^2+{(\gamma \omega_r)}^2}} [/tex]


    3. The attempt at a solution

    I have subbed in the equation for the amplitude into the equation for the power and then differentiated it and let it equal to zero to find the max power. The differentiated gets pretty messy but I still cant seem to get to the right answer that [itex] \omega_r = \omega_0 [/itex] no matter how many times I differentiate it. My notes from lectures has the proof but my lecturer has differentiated the equation in a really unusual way so I cant follow it.
     
  2. jcsd
  3. Jan 12, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    Try dividing numerator and denominator of ##\bar{P}## by ##\omega_{r}^2## and simplifying. I don't think you'll need calculus to see the result.
     
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