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Work and Power of the Friction Force in an F=-bυ damped oscillation

  1. Apr 17, 2010 #1
    Hey there forum!

    Consider a damped oscillation in which the friction force is F=-bυ.
    What I want to ask is how do you calculate the work done by this force for any x interval along a line and what is the Power of the work done by this force?

    I already know that Power P of the work done due to a force F is P=Fυ, therefore substituting would give P=-bυ2. But I cannot derive the equation P=Fυ from my equations.

    Thanks in advance!
  2. jcsd
  3. Apr 18, 2010 #2


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    Hey there karkas! :smile:
    Fv = force times speed = force times distance per time = work done per time = energy per time = power = P :wink:
  4. Apr 18, 2010 #3
    Yea, thanks tiny-tim ! I got that solved too, but I'm kinda still stuck as to calculating the work done by the force F=-bυ, say during T/2 of the oscillation.

    All I know is that [itex]W=\int_{0}^{A}Fds[/itex] but can't work on it. Doesn't really matter though, since the question that I wabted to answer is answered now!
  5. Apr 18, 2010 #4

    Andy Resnick

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    Perhaps it's a minor point, but the frictional force does not do work. It's better to do these calculations from the energy perspective. For a damped oscillator:


    You can calculate v(t) and from that, the kinetic energy. The difference between the kinetic energy at time t and time '0' is the cumulative amount of dissipated energy from friction. You can also calculate this in terms of the power (energy * time) if you wish.
  6. Apr 18, 2010 #5
    Boy, you're a genius! Thanks for changing my perspective, that's hella much wiser. But I think that in a damped oscillation (the underdamped one) there is a work of the frictional force and it expresses the energy lost by the system.

    So in such an oscillation, where [itex]A=A_0 e^{-kt}, k=b/2m[/itex] the work done by the frictional forces equals to [itex]W= E - E_0 = 1/2 DA^2 - 1/2DA_0^2[/itex] right?
    Last edited: Apr 18, 2010
  7. Apr 18, 2010 #6

    Andy Resnick

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    I'm not sure I'm following you, but for the underdamped case where x(t) = x_0 exp (-ct/2m)cos (w_d t), the velocity is then v(t) = x_0 * c/2m * w_d * exp (-ct/2m) sin(w_d t) and the frictional loss of energy g(t)= 1/2m(v_0^2-v(t)^2) which may reduce to your result every 1/2 period.

    Edit- this is true for the restrcited case x = 0. Otherwise, the potential energy of the spring [1/2 kx(t)^2] has to be included.

    In any case, I'm glad to be of help!
    Last edited: Apr 18, 2010
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