# Power Dissipation

1. Sep 8, 2009

### swuster

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

2. Relevant equations
n/a

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

$$\omega/2\pi*\int^{2\pi/\omega}_{0} -bv_x^2 dt$$

But what is $$v_x$$? If $$x(t) = \underline{A} e^{i \omega t}$$, then $$v(t) = i \omega \underline{A} e^{i \omega t} = i\omega x(t)$$. So then I think that $$v_x = i\omega$$ but this doesn't give me the correct answer when put into the integral. Thanks for the help!

Last edited: Sep 9, 2009
2. Sep 9, 2009

### ehild

Use the real solution,

$$x(t) = \underline{A} cos(\omega t)$$ or $$\underline{A} sin(\omega t)$$.

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.

ehild

3. Sep 9, 2009

### swuster

So then if I just use $$x(t) = A cos(\omega t)$$, then it follows that $$v(t) = -\omega A sin(\omega t)$$. Is $$v_x$$ just dv/dt / dx/dt then? That would make it $$\omega cot(\omega t)$$ which also does not work in the integral

Last edited: Sep 9, 2009
4. Sep 9, 2009

### ehild

I do not understand what you are doing. $$v_x$$ 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, $$v_x = dx/dt$$ .

ehild