Classical Mechanics: Lightly Damped Oscillator Driven Near Resonance

  • Thread starter RylonMcknz
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  • #1
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!
 

Answers and Replies

  • #2
vanhees71
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You are on the right track! Just do the integral. It might help to rewrite the sine in terms of exps!
 
  • #3
Should I expand sine as a series or use the identity sin(z)=[e^(iz)-e^(-iz)]/2i?
 
  • #4
AlephZero
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Use the identity sin(z)=[e^(iz)-e^(-iz)]/2i.

Or better still, take the general solution as the real part of ##Ce^{st}## where ##s = -b + i\omega## and ##C## is a complex constant.
 
  • #5
78
12
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!


As a sanity check, you can work the problem again with Q being defined as reactance over resistance. You should get the same answer. :)
 
  • #6
Thank you everyone for the help. I've been working the problem when I have time. When I put sine in the exp form and integrate, it looks terrible. But a further hint from the professor says that I should be looking for ΔE/E. I think that is the key, but I'm still unsure. I'll be able to work it further when I get home. Again, thanks so much for the help.
 
  • #7
I'm still unable to solve the problem. I can't seem to make this integral manageable enough to continue. I attempted to use Wolfram Alpha to integrate before and after putting sine in exp form, but I'm still unable to move forward. I feel like there must be some terms I should be neglecting because of the light damping.
 
  • #8
78
12
Try this:

Determine the energy dissipation per cycle. (I know, this is cheating, but you'll get some great insights!)

Eric
 
  • #9
vanhees71
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I still don't see, where there might be a problem. You have an integral of the form
[tex]\int \mathrm{d} t \exp(-a t) \sin^2(b t)[/tex]
with constants [itex]a[/itex] and [itex]b[/itex]. Now you write
[tex]\sin(b t)=\frac{\exp(\mathrm{i} b t)-\exp(-\mathrm{i} b t)}{2 \mathrm{i}},[/tex]
take the square and multiply it out. Then you end with integrals of the form
[tex]\int \mathrm{d} t \exp(c t),[/tex]
and over a constant which are really easy to deal with.
 

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