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Resistive Force (of a damped oscillator: what is it?)

  1. Jan 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a damped oscillator, with natural frequency [tex]\omega_{o}[/tex] and damping constant B, both fixed, that is driven by force F(t) = F[tex]_{o}[/tex]cos([tex]\omega[/tex]t).


    Verify that the average rate at which energy is lost to the resistive force is mB[tex]\omega^2[/tex]A[tex]^2[/tex].

    2. Relevant equations
    x = Acos([tex]\omega[/tex]t-[tex]\delta[/tex])
    v = -A[tex]\omega[/tex]sin([tex]\omega[/tex]t-[tex]\delta[/tex])
    Fr = -bv
    b = 2Bm
    T = period = (2pi)/omega

    3. The attempt at a solution
    The resistive force is Fr = -bv, ehe?

    Where b = 2Bm?

    So I'll say P(t) = (Fr)(v)

    So then the average rate <P> at which the energy is lost to the resistive force is

    <P> = (1/T) times the integral(from zero to T) of P(t)dt
    = ((-2BmA[tex]\omega[/tex])/T) times the integral(from zero to T) of sin([tex]\omega[/tex]t-[tex]\delta[/tex])

    But if I do that I obviously get zero.

    A solution I saw had what I have, except the velocity was squared. I don't know why the velocity had been squared. Please, please, can someone explain to me why the velocity is squared? That would be lovely. Very, very, oh so very and extremely lovely.



    Also, I don't know why some of the omegas are high. Sorry.
     
  2. jcsd
  3. Jan 29, 2010 #2

    vela

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    The power is equal to force times velocity, and the force is proportional to the velocity. So power is proportional to v2.
     
  4. Jan 29, 2010 #3
    Ya....

    Thank you. It was 7am when I posted this, and I hadn't slept. I realized my mistake as I thought about this problem in the shower.
     
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