Imagine a SHO with a drive F(t). (or in general a Hamiltonian system)

What is the power delivered to the system and can we talk about the power reflected? is i am imagining say a MW oscillator cavity fed by a transmission line.

My 3 attempts at the question:

1) One definition of power I know is $P = \dot{U}$, where $U=\int F(x,t) \cdot dx(t)$ is the energy. so that the power delivered $\dot{U}=\frac{d}{dt} \int F(x,t)\cdot dx(t)$ which is a funcky chair rule derivative?? This is where I get stuck on #1

2) Another definition for power i know is $P = F(x,t)\cdot v(t)=F\cdot\frac{\partial \mathbb{H}}{\partial p}$
This may also explain how to do the "chain rule in #1": ie $\dot{U}=\frac{d}{dt} \int F(x,t)\cdot dx(t)=\frac{d}{dt} \int F(x,t)\cdot \frac{dx(t)}{dt}dt=\frac{d}{dt} \int F(x,t)\cdot v(t) dt=F(x,t)\cdot v(t)$ so maybe it is this simple?

Where does this definition of Power come from / why would it be more fundamental?

3) Another approach I thought of is take the Hamiltonian term of the drive: $\mathbb{H} $ ~ $F(t)x $ and the power on this energy, should be: $P=\dot{U} = \frac{d}{dt} (F(t)\cdot x(t))=F\cdot v + \dot{F}x$ So here there are TWO terms! the power from part (2) $F\cdot v$ and a new term, which I cannot account for: $\dot{F}x$ What is this term and why doesn't this approach work? Is this the reflected power?

I am sorry for this basic question, but I hope it will help clear up some fundamental things. :)

  • $\begingroup$ I can think of at least two general ways to approach this. The one I would use depends on the context. If it is a mechanical oscillator (like a pendulum), your description seems like the natural approach. But if it is an EM resonator cavity (as your second paragraph suggests), then I would try something like integrating the Q of the cavity. Can you clarify what the system actually is? $\endgroup$ – Colin McFaul May 31 '13 at 14:35
  • $\begingroup$ Does Q refer to charge? Yes on some deep level the system is a MW cavity, but I am only thinking of it near resonance so it is simplified to an RLC parallel circuit, which is formally equivalent to a damped driven mass-spring system. I hope this clarifies? I am not sure what you mean by integrate the Q of the cavity $\endgroup$ – AimForClarity May 31 '13 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.