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**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:**

I have spent much time attempting to figure this out. I think that the frictional force is

Any help would be greatly appreciated.

Thank you!

**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!