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

1. Apr 17, 2010

### karkas

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.

2. Apr 18, 2010

### tiny-tim

Hey there karkas!
Fv = force times speed = force times distance per time = work done per time = energy per time = power = P

3. Apr 18, 2010

### karkas

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 $W=\int_{0}^{A}Fds$ but can't work on it. Doesn't really matter though, since the question that I wabted to answer is answered now!

4. Apr 18, 2010

### Andy Resnick

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:

http://en.wikipedia.org/wiki/Damping

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.

5. Apr 18, 2010

### karkas

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 $A=A_0 e^{-kt}, k=b/2m$ the work done by the frictional forces equals to $W= E - E_0 = 1/2 DA^2 - 1/2DA_0^2$ right?

Last edited: Apr 18, 2010
6. Apr 18, 2010

### Andy Resnick

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