Undergrad What's the interpretation of the third term F(t)q?

[Raffealla]

TL;DR

I have calculated the rest questions. I guess it's maybe a special kind of potential energy, but it's seems like it's different from what the following questions suggests.

I think that time derivative of the energy of the oscillator is F times the derivative of q, which means it's the power of the external force. So it's like it is suggesting that F·q is the work done by external force, which makes no sense at all. As far as I concerned, the input energy has no relation with current displacement. What's the interpretation of the third term Fq? And if my calculation is correct ,time derivative of Hamiltonian is the derivative of F times q. Why is the H different from E? What does the H represent here?

 

[anuttarasammyak]

F(t)=-\frac{\partial U_{ex}}{\partial q}
where
U_{ex}(t,q)=-F(t)q
potential energy of external force. Hamiltonian is
H=T+U+U_{ex}

 

[TSny]

I think that time derivative of the energy of the oscillator is F times the derivative of q, which means it's the power of the external force.

Yes.

So it's like it is suggesting that F·q is the work done by external force, which makes no sense at all. As far as I concerned, the input energy has no relation with current displacement. What's the interpretation of the third term Fq?

I don't see a "physical" interpretation of $F(t) q$. I agree that $F(t) q$ does not represent the work done on the oscillator by the external force $F(t)$. Of course, $F(t) q$ is necessary so that $F(t)$ will appear as the driving force in the equations of motion.

And if my calculation is correct ,time derivative of Hamiltonian is the derivative of F times q.

Ok. But I get a negative sign $\dot H = -\dot F q$.

Why is the H different from E? What does the H represent here?

For general dynamical systems, $H$ does not necessarily equal the energy of the system. There are systems for which $L = T - V$ but the Hamiltonian does not have the form $H = T+V$. For example, this can occur if you have a velocity-dependent potential; i.e., $V$ is a function of both $q$ and $\dot q$.

For this problem, you could write $L = T-\widetilde{V}(q, t)$, where $\widetilde{V}(q, t) \equiv V_{\rm osc}(q) + F(t)q = \frac 1 2 \omega^2 q^2 + F(t) q$. Here, $\widetilde V$ does not depend on the velocity $\dot q$ and so it's not surprising that we find $H(p, q) = T + \widetilde V$. Here, $T$ is the kinetic energy expressed in terms of the momentum. (For this problem, the momentum and the velocity are the same: $p = \dot q$.)

So, the Hamiltonian looks like it expresses some total energy. But it's not the total energy of the oscillator. Instead, $H = E_{\rm osc} -F(t)q$. I don't see a nice physical interpretation of the Hamiltonian in terms of energy for this problem.