Hello Physics Forum! I have a question: The problem: For a lightly damped oscillator being driven near resonance in the steady state, show that the fraction of its energy that is lost per cycle can be approximated by a constant (something like 2pi, which is to be determined) divided by the Q factor (Q is defined as the resonant frequency of a driven damped oscillator divided by 2*β, where β is the damping parameter). My professor gave this hint to get us started: ΔE = ∫Fdx. Where E is the energy lost, and F is the force of friction. We are supposed to integrate from 0 to τ(cycle/period), and the professor suggested to change dx to (dx/dt)dt = vdt. I have spent much time attempting to figure this out. I think that the frictional force is F=-bv, where b is some positive constant and v is the velocity. I try to use the solution of the differential equation for such motion, which is x(t)=Ae^(-βt)cos(ωt-δ). I take the derivative of this to get v(t). The second term in v(t) can be ignored because the damping is light. So I have: ΔE = -b∫v^2dt from 0 to τ, where v(t)≈-Ae^(-βt)ωsin(ωt-δ). This integral makes a nasty mess that doesn't get me anything useful. I think I need to simplify this further by approximation, but I don't know which assumptions to make. Any help would be greatly appreciated. Thank you!