Question about a coupled system of pendulums

[LCSphysicist]

Homework Statement

All below

Relevant Equations

There is no.

The question is:
We have a system with two pendulum attached to a spring, both pendulum are connected in the ceiling by strings. The strings, and the bobs are equal.
So now you pick the bob A and pull it to an initial displacement, maintaining the second bob in rest. Now, you let the first go. So the initial conditions are:

Xo(A) = Xa
Xo(B) = Vo(A) = Vo(B) = 0

You see the system made one oscillation, and you now go out the room.
After a time, you returned in the exact moment the pendulums have the same energy, the same displacement and so.

The time stop, the question is: How would you know which pendulum will lose energy, which will gain or if the energy of both pendulums will not change.

I am not sure if i get the question...
At first i would say it is impossible to give a answer, we can not distinguish. But i think if we know the phases in this instant, we can give and answer, so my answer would be:

"If we know, by some reason, what are the phase in this instant, we can give an answer (with phase i want to say: if we know the pendulum A is, let's say, one quarter of period after the start). Otherwise, it is not possible"

What do you think?

 

[kuruman]

The question itself, "which pendulum will lose energy" does not make much sense to me assuming that you are talking about mechanical energy, not just kinetic. A pendulum might lose kinetic energy if the bob is moving slower at $t_2$ than at $t_1$, but it does not "have" its own potential energy to lose because it shares that with other entities. There is gravitational potential energy $U_{EA}$ of the Earth-pendulum A system, gravitational potential energy $U_{EB}$ of the Earth-pendulum B system and elastic energy $U_{AB}$ stored in the spring that connects the two. We know that the sum of all the potential and kinetic energies is always the same, but other than that, I wouldn't know how to split this sum into two pieces, one for each pendulum.

Of course, if you are concerned with kinetic energy only, the answer is easy: none of the two will lose kinetic energy when released because they are both at rest.

 

[LCSphysicist]

The question itself, "which pendulum will lose energy" does not make much sense to me assuming that you are talking about mechanical energy, not just kinetic. A pendulum might lose kinetic energy if the bob is moving slower at $t_2$ than at $t_1$, but it does not "have" its own potential energy to lose because it shares that with other entities. There is gravitational potential energy $U_{EA}$ of the Earth-pendulum A system, gravitational potential energy $U_{EB}$ of the Earth-pendulum B system and elastic energy $U_{AB}$ stored in the spring that connects the two. We know that the sum of all the potential and kinetic energies is always the same, but other than that, I wouldn't know how to split this sum into two pieces, one for each pendulum.

Of course, if you are concerned with kinetic energy only, the answer is easy: none of the two will lose kinetic energy when released because they are both at rest.

Oh yes, i made a flaw, the most correct word here is increase and decrease. There is another way to question too:
"How do the system know what way the energy should flow?"
Remembering this is classical physics, if there is a way to know by statistical or quantum that i don't know, please try to don't use it ;C

 

[hutchphd]

I do not understand the question. What does

what way the energy should flow

mean? What energy? Flow where?
The initial system energy will slowly flow out of the system to external degrees of freedom if allowed ...The remaining energy will cycle through the various parts of the system. The sum of all will be constant.

 

[jbriggs444]

I am not sure if i get the question...

Nor am I sure that I get the question either. But I have a theory.

Before we leave the room, we have set up initial conditions for displacement and velocity of everything. We watch the system evolve for a while. Still nice and deterministic. But then we leave.

The problem states that we leave after "one oscillation". But it is a complex system. It will not have a simple periodic behavior. I do not believe that "one oscillation" means that the system has returned to its initial state before it was released. I think it just means that we leave. The pendulum bobs are not necessarily at rest.

When we return we are told that the pendulum bobs have the same positions and energy as when we left. But we do not know which directions they are moving. I think that the question is a back-handed way of getting us to worry about that.

 

[hutchphd]

Aaah "same" means same as t=0 not "equal". Thanks, I believe that is the intent.

 

[TSny]

Hi, @LCSphysicist. Because of the confusion in interpretation of the question, can you please post the exact wording for the entire question?

 

[hutchphd]

At any subsequent time one needs two initial conditions to specify subsequent motion. I believe that is equally true if time propagates backwards, so I believe the answer is, given only the positions , there are no way to distinguish the "direction" for an isolated system.

 

[kuruman]

But wait a minute! We have observed which way the system will move when we first released it. At the moment of release all the mechanical energy is potential because we are told that both bobs start from rest. So when we return and observe the bobs at the same positions, again all the energy must be potential and the bobs will again be (instantaneously) at rest with zero kinetic energy. Therefore, immediately after that the bobs will move the same way as when we first released the system.

 

[jbriggs444]

But wait a minute! We have observed which way the system will move when we first released it. At the moment of release all the mechanical energy is potential because we are told that both bobs start from rest. So when we return and observe the bobs at the same positions, again all the energy must be potential and the bobs will again be (instantaneously) at rest with zero kinetic energy. Therefore, immediately after that the bobs will move the same way as when we first released the system.

We do not leave immediately after releasing it. We leave after "one oscillation". If the system has a simple period, you are correct, the system would then have returned to a state of rest.

But if the behavior is more complicated, everything might not be back at rest. That is the interpretation that I think makes sense. But I agree that it is certainly not made clear.

Edit: It is also conceivable that the system is aperiodic -- that there are no repeating states and that we can never come back. But the problem statement does not contemplate such a scenario.

 

[kuruman]

But if the behavior is more complicated, everything might not be back at rest. That is the interpretation that I think makes sense. But I agree that it is certainly not made clear.

I agree that the motion is complicated, but I would argue that when we return and see the bobs in the same configuration they must be instantaneously at rest. I have laid out the arguments below. If you find that one is incorrect or that a conclusion is unjustified, please point that out.
1. The potential energy of the system depends only on the position of the bobs relative to each other and relative to the Earth.
2. At the moment of release with the bobs at their initial positions, their kinetic energy is zero because they are released from rest. Therefore, the mechanical energy of the system is initially entirely potential.
3. Mechanical energy is conserved and is equal to the initial potential energy at all times.
4. When we return and see the bobs in their initial positions, we conclude that they have the same potential energy as initially. This leaves zero kinetic energy for the bobs which means that they are not moving (instantaneously) when we return.
5. A moment later the system will evolve exactly the same way as when it was under observation for the first time because the initial conditions are the same.

 

[jbriggs444]

I agree that the motion is complicated, but I would argue that when we return and see the bobs in the same configuration they must be instantaneously at rest. I have laid out the arguments below. If you find that one is incorrect or that a conclusion is unjustified, please point that out.
1. The potential energy of the system depends only on the position of the bobs relative to each other and relative to the Earth.
2. At the moment of release with the bobs at their initial positions, their kinetic energy is zero because they are released from rest. Therefore, the mechanical energy of the system is initially entirely potential.
3. Mechanical energy is conserved and is equal to the initial potential energy at all times.
4. When we return and see the bobs in their initial positions, we conclude that they have the same potential energy as initially. This leaves zero kinetic energy for the bobs which means that they are not moving (instantaneously) when we return.
5. A moment later the system will evolve exactly the same way as when it was under observation for the first time because the initial conditions are the same.

We do not leave at the moment of release. We leave "one oscillation" later. It is not completely clear that everything is at rest when we leave.

 

[LCSphysicist]

HI.
In #1 i posted my interpretation too, in a way that explain what is the crucial problem of the question, but I think i overestimated the issue

Actually, maybe the main problem here is that we are not necessarily to go out of the room, i read again and see that we follow the motion one oscillation entirely, and so until they have the same energy. I interpreted we follow the motion one oscillation entirely, and returned to see when they have the same energy.
(the second probably allow us to use the difference of phases, the second not allow it)

But is good to know the second maybe is impossible to know, it helps me too.
Sorry.

 

[kuruman]

We do not leave at the moment of release. We leave "one oscillation" later. It is not completely clear that everything is at rest when we leave.

Agreed. However, we are there at the moment of release and also there when the system returns to the same configuration it had at the moment of release. Therefore, the time interval between the moment of release and the time we return must be an integer multiple of the period, defined as the time required for the system to return to its initial conditions. So when we return we have a good idea of how the system will evolve until a time equal to "one oscillation" has elapsed after our return.

 

[jbriggs444]

Agreed. However, we are there at the moment of release and also there when the system returns to the same configuration it had at the moment of release. Therefore, the time interval between the moment of release and the time we return must be an integer multiple of the period, defined as the time required for the system to return to its initial conditions. So when we return we have a good idea of how the system will evolve until a time equal to "one oscillation" has elapsed after our return.

If one interprets "one oscillation" to be one period of the system state then I agree completely.

 

[kuruman]

If one interprets "one oscillation" to be one period of the system state then I agree completely.

It could be that, but it doesn't have to be.

You see the system made one oscillation, and you now go out the room.
After a time, you returned in the exact moment the pendulums have the same energy, the same displacement and so.

The time interval from starting the system until we return is $\Delta t =$ "one oscillation" + "after a time". Each of the terms in the sum could be anything, but the sum should be equal to an integral multiple of the period.

 

[hutchphd]

Yes if we are told that the system has that initial configuration the velocities must be zero and it is groundhog day (movie allusion) all over again!
But an arbitrary system may never make it back to the initial state(?the issues of commensurate frequencies...at least in finite time?).
I hesitate to ask what if we put in the nonlinearities into the pendula. Because they are identical can an argument still be crafted?
Food for thought.