I Damped oscillator with changing mass

[Orodruin]

Also, if you let the exhaust velocity depend on pressure at the bottom ... Then the effective $g$ is going to depend on the instantaneous acceleration of the mass - it will not be constant. The rate of mass loss is also going to depend on this etc etc. This is why things get complicated fast.

 

[Delta2]

Also, if you let the exhaust velocity depend on pressure at the bottom ... Then the effective $g$ is going to depend on the instantaneous acceleration of the mass - it will not be constant. The rate of mass loss is also going to depend on this etc etc. This is why things get complicated fast.

By effective g you mean the g that the water "sees", i.e the g that we use to calculate the exhaust velocity of water?

 

[Orodruin]

By effective g you mean the g that the water "sees", i.e the g that we use to calculate the exhaust velocity of water?

Yes, since the mass is accelerating, this will lead to an effective value of g different from the pure gravitational acceleration.

 

[alan123hk]

@alan123hk I think you have a critical mistake , in the rocket equation ve is the relative velocity of exhaust gas to the rocket, so it shouldn't contain v but only the first term.

Yes, since the mass is accelerating, this will lead to an effective value of g different from the pure gravitational acceleration.

Thanks for pointing out where the derivation went wrong.
So is the correct expression $V_e=\sqrt{2~(g+\frac{dv}{dt})~h} ~~$?
Are there any other mistakes ?

Thanks.

 

[tech99]

You are altering one element of the oscillator in a linear manner; it is similar to the parametric oscillator, although in this case we usually alter one of the elements cyclically. https://en.wikipedia.org/wiki/Parametric_oscillator

 

[Rezex124]

Okay so today I went to my prof., and I showed him how I approached the problem. We derived the same equations we got here together, and I asked him how he would go on to get the oscillation time. He said that I definitely need to solve this numerically, as someone already suggested.

He advised that I can use a program called Berkeley Madonna, and that I model a flowchart with my data, and how they interact with each other (e.g. mass with velocity etc.). Then to plot a graph, on which there would be both oscillations (constant vs. non-constant mass, at same initial conditions), and compare the periods, oscillation times, displacement and so on. Then to ''subtract'' the graphs and see where the main differences are.

Basically I shouldn't worry how to get a generalized equation, but to focus on the oscillation time for my specific experiment to get a solution.And again, thank you all for your contributions :)

 

[haruspex]

Then to ''subtract'' the graphs and see where the main differences are.

I doubt simply subtracting them will reveal much. The changing mass will change the time base. I would suggest instead:
- solve as though mass were constant
- plug in the mass as a function of time, as previously suggested, so as to get a changing frequency
- then do the subtraction to see how well that serves as an approximation
Of course, it might happen to look ok with some parameters but fail badly with others. Near resonance would be an interesting test.

 

[Rezex124]

Will try that for sure sometime next week. Got more pressing stuff to deal with for now.

Thanks for the suggestion

 

[bob012345]

When we consider a system where the total mass is variable ,Newton's 2nd law doesn't hold. It would violate conservation of energy and conservation of momentum if it hold and @haruspex example shows that clearly. Newton's 2nd holds only if we consider a system where the total mass is constant. But in that case depending on the system we might not be able to write the total momentum $p$ as $p=mv(t)$ where m the total constant mass of the system , because different parts of the system might have different velocities and such is the case with a rocket and also the case here.

In the differential form the mass is not assumed constant so I believe Newton's Second law as Newton formulated it always holds you just have to be careful.

 

[Delta2]

In the differential form the mass is not assumed constant so I believe Newton's Second law as Newton formulated it always holds you just have to be careful.

No I disagree it doesn't holds for systems where the total mass of system is variable. Take the example of @haruspex with a cart that is dripping water. If we take as a system only the cart and not the water that escapes , then if we apply Newton's second law we ll find (due to the term $\dot m v$, $F=0\Rightarrow m\frac{dv}{dt}+\frac{dm}{dt}v=0$) that the cart is accelerating, but the water is just dripping so it doesn't carry additional momentum, so where does the extra momentum of the cart is coming from. We have violation of conservation of momentum and also of conservation of energy.

If we consider as system the cart + the water that is dripping then the law holds but then the total mass of this system is constant.

 

[bob012345]

No I disagree it doesn't holds for systems where the total mass of system is variable. Take the example of @haruspex with a cart that is dripping water. If we take as a system only the cart and not the water that escapes , then if we apply Newton's second law we ll find (due to the term $\dot m v$, $F=0\Rightarrow m\frac{dv}{dt}+\frac{dm}{dt}v=0$) that the cart is accelerating, but the water is just dripping so it doesn't carry additional momentum, so where does the extra momentum of the cart is coming from. We have violation of conservation of momentum and also of conservation of energy.

If we consider as system the cart + the water that is dripping then the law holds but then the total mass of this system is constant.

Maybe it's just semantics but what I hear you saying is if you misapply NSL it doesn't work so NSL doesn't always work. That seems an entirely confusing and unecessary approach guaranteed to confuse students. Better to say it always applies but you just have to be careful how things are defined.

I very much doubt Newton himself would say his law doesn't apply in certain cases.

 

[Delta2]

Maybe it's just semantics but what I hear you saying is if you misapply NSL it doesn't work so NSL doesn't always work. That seems an entirely confusing and unecessary approach guaranteed to confuse students. Better to say it always applies but you just have to be careful how things are defined.

I very much doubt Newton himself would say his law doesn't apply in certain cases.

Ok since you think I misapply NSL, tell me how YOU correctly apply it for the cart that is dripping water (not for the whole system cart+escaped water but only for the cart).

 

[erobz]

Ok since you think I misapply NSL, tell me how YOU correctly apply it for the cart that is dripping water (not for the whole system cart+escaped water but only for the cart).

Don't you still have to choose a direction? In the case of the cart dripping water from the bottom, the total momentum ( cart + drop) in the direction of motion remains constant. That is to say the mass flowrate in the direction of motion is zero. Sure, the cart is losing mass, but its not going to change velocity.

For the draining cup the mass flowrate in the direction of motion is non-zero, and there will be an impulse in the direction of motion from the fluid jet.

And something seems to be mis applied in taking

$$ \sum F = \frac{dM}{dt} v + M \frac{dv}{dt} $$

The Rocket Equation is derived from impulse in a stationary frame of reference $O$ :

$$ \sum F = \frac{dM}{dt} v_{e/R} + M \frac{dv}{dt} $$

The $v$ in the first would be with respect to $O$: i.e. $v = v_{e/O}$

 

[Delta2]

For the draining cup the mass flowrate in the direction of motion is non-zero, and there will be an impulse in the direction of motion from the fluid jet.

So you think the term $\frac{dm}{dt}\frac{dx}{dt}$ is correct in equation of post #2? I just don't think so but I will come back on this tomorrow, now I got to go and try to sleep.

 

[erobz]

So you think the term $\frac{dm}{dt}\frac{dx}{dt}$ is correct in equation of post #2? I just don't think so but I will come back on this tomorrow, now I got to go and try to sleep.

No, not exactly. There should ( or at least could ) be a thrust term, but not quite that. For this following example, its probably very small in comparison with the weight.

 

[erobz]

For the draining cup $c$ hanging from a simple string with tension $T$ I don't see why the rocket equation wouldn't apply, but that doesn't mean that I'm not missing something. The oscillating cup has to be a completely different animal, but I think maybe start here?

Let $M$ be the total mass

$$ M = m_c + m_w $$

Then

$$ \sum F = \frac{dM}{dt} v_{e/c} + M \frac{dv}{dt} = T - Mg $$

Without viscous effects , where $v_{e/c}$ is the velocity of the jet relative to the cup:

$$ v_{e/c} = \sqrt{2gz}$$

where $z$ is the height if the water in the cup

Also we see that:

$$ m_w = \rho A_c z $$

Where $\rho$ is the density of water, and $A_c$ is the cross-sectional area of the cup.

Thus;

$$ v_{e/c} = \sqrt{ \frac{2g m_w}{ \rho A_c}} $$

It also follows that:

$$ \frac{dM}{dt} = \frac{d m_w}{dt} = -\rho A_h v_{e/c} = - \rho A_h \sqrt{ \frac{2g m_w}{ \rho A_c}} $$

Where $A_h$ is the area of the drain hole.

Putting that all together I get:

$$ - \rho A_h \sqrt{ \frac{2g m_w}{ \rho A_c}} \sqrt{ \frac{2g m_w}{ \rho A_c}} + M \frac{dv}{dt} = T- Mg $$

$$ -2 g \frac{A_h}{A_c} m_w + M \frac{dv}{dt} = T - Mg $$

I'm pretty sure that in this particular example - a tank draining under gravity - is not going to "take off" i.e. $\frac{dv}{dt} = 0$, and the tension force in the string is just less than the total weight of the cup and remaining water.

$$ T = Mg - 2 g \frac{A_h}{A_c} m_w $$

Does this seem reasonable?

 

[bob012345]

Ok since you think I misapply NSL, tell me how YOU correctly apply it for the cart that is dripping water (not for the whole system cart+escaped water but only for the cart).

As I said it may just be semantics and I am not saying you are wrong or misapplying it but what I said is what I thought I heard you were saying at least as it strikes me. I will try and resolve it.

First, let's look at Newton's law. Newton's Second Law of Motion in 1792 english reads:

"LAW II: The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed. � If a force generates a motion, a double force will generate double the motion, a triple force triple the motion, whether that force be impressed altogether and at once, or gradually and successively. And this motion (being always directed the same way with the generating force), if the body moved before, is added to or subtracted from the former motion, according as they directly conspire with or are directly contrary to each other; or obliquely joined, when they are oblique, so as to produce a new motion compounded from the determination of both."

The way it reads to me is that if there is a force it causes a change of momentum which is subtly different from saying if there is a change in momentum there must be a corresponding force which is what we are debating here.

If an asteroid is careening towards Earth and breaks into multiple pieces along the way we don't say there is a force on it changing its motion in terms of speed or direction because the momentum of the central piece is lower after parts separate. Of course if some force such as expanding gasses acts between two parts of the asteroid and causes them to separate then that force will lead to a change in momentum of both parts in the direction of the net force on each.

So applying this to the cart dripping water we see there is no force and therefore no change in momentum of the cart due to an impressed force as Newton says it along the line of motion. In this case there is no net force along the line of motion so how do we apply LAWII above? The law states that the alteration of motion is ever proportional to the motive force impressed which is zero so borrowing from you @Delta2

$$F=0\Rightarrow m\frac{dv}{dt}+\frac{dm}{dt}v=0$$

The way I view how LAWII applies is not that the sum of these terms is zero thus making a spurious force but that both terms are zero. The $\dot{m}$ applies to a momentum change caused by an net impressed force along the line of motion and not the change in mass of the cart due to dripping water. The Second Law holds always but has no effect here. The Second Law is true at every instant before, during and after every drop and as I can't resist, you might say it's good to the last drop..
https://www.physics.utoronto.ca/~jharlow/teaching/everyday06/reading01.htm

 

[bob012345]

Summary: I want to study how the oscillation time changes, if I compare an oscillator with constant vs. non constant mass
...

I would kindly ask, if someone can nudge me into a right direction, a good source or a possible solution.
My only goal right now is to get the equation of oscillation time for damped oscillation with changing mass.Thank you in advance,
Tine

See if these papers help;

Sand;
https://www.researchgate.net/publication/228412656_Variable_mass_oscillator/link/00b4951ad53e60ac25000000/download

Water;
http://www.scielo.org.mx/pdf/rmfe/v60n1/v60n1a3.pdf

 

[erobz]

See if these papers help;

Sand;
https://www.researchgate.net/publication/228412656_Variable_mass_oscillator/link/00b4951ad53e60ac25000000/download

Water;
http://www.scielo.org.mx/pdf/rmfe/v60n1/v60n1a3.pdf

Awe man...the fun police arrived

In all honesty, I'm glad to see I wasn't on a path to nowhere.

 

[bob012345]

Awe man...the fun police arrived

I only referenced them so the OP can have them to compare with not that the problem has to be addressed exactly like these. It doesn't even look like these two do the same treatment anyway so I think there is still room for fun.

 

[erobz]

I only referenced them so the OP can have them to compare with not that the problem has to be addressed exactly like these. It doesn't even look like these two do the same treatment anyway so I think there is still room for fun.

For the water bucket on a spring: I'm kind of sad they used Bernoulli's without mentioning that it is derived in an inertial frame. In the frame of reference of the cup, the pressure at the nozzle will oscillate around $\rho gz$. It will inevitably be a function of $\frac{dv}{dt}$ and $z$ itself. I would have liked to see how they neglected it professionally. I would say they hand waived it away, but actually they didn't address that at all. I'm sure it's a fine assumption for all practical purposes, but I would have liked to see it regardless for my own future reference.

 

[Rezex124]

See if these papers help;

Sand;
https://www.researchgate.net/publication/228412656_Variable_mass_oscillator/link/00b4951ad53e60ac25000000/download

Water;
http://www.scielo.org.mx/pdf/rmfe/v60n1/v60n1a3.pdf

I just had a quick glance over them and yes! They will definitely help.

Thank you very much

 

[bob012345]

I just had a quick glance over them and yes! They will definitely help.

Thank you very much

You're welcome but remember just because these papers were published does not mean they are the last word and the best way or only way to do this. Maybe you and your teacher can put together your own approach and publish it.

 

[erobz]

For the water bucket on a spring: I'm kind of sad they used Bernoulli's without mentioning that it is derived in an inertial frame. In the frame of reference of the cup, the pressure at the nozzle will oscillate around $\rho gz$. It will inevitably be a function of $\frac{dv}{dt}$ and $z$ itself. I would have liked to see how they neglected it professionally. I would say they hand waived it away, but actually they didn't address that at all. I'm sure it's a fine assumption for all practical purposes, but I would have liked to see it regardless for my own future reference.

Never mind, I see that they do mention this issue:

"At this point, we point out that the quadratic dependence of mass on time is only a motivator for the leaking oscillator problem, treated here as a purely theoretical problem. Therefore, this result should be considered within its appropriate limitations. Probably, when the bucket is moving, going up and down with the oscillations, the flow rate through the hole could be seen to change as well, deviating slightly from the results obtained here. In other words, we are ignoring the fact that the bucket, as well as the water within it, are accelerating frames. However, we can admit that the quadratic dependence of mass must work as a reasonable approximation in the case the loss of water occurs at a very low rate, and the oscillating bucket experiences smooth motions as investigated in this article"[1]

[1] http://www.scielo.org.mx/pdf/rmfe/v60n1/v60n1a3.pdf

 

[Delta2]

The way I view how LAWII applies is not that the sum of these terms is zero thus making a spurious force but that both terms are zero.

How the term $\dot m v$ can be zero (assuming that the cart has initial velocity)?. And yes we apply the law in the direction of $\vec{v}$, not in the direction of dripping where there is external force (the weight ). And no you just can't say $\dot m$ is zero because the force that causes it is perpendicular to the direction we apply the law, no sorry if the mass is varying w.r.t time then $\dot m\neq 0$ simple as that...

 

[Delta2]

I don't see why the rocket equation wouldn't apply

The rocket equation applies but this equation is derived by applying Newton's 2nd law in a system where the total mass is constant that is the rocket+the exhaust gases.

 

[bob012345]

How the term $\dot m v$ can be zero (assuming that the cart has initial velocity)?. And yes we apply the law in the direction of $\vec{v}$, not in the direction of dripping where there is external force (the weight ).

Because there is no force creating a momentum change. The second law exists and holds, but has a null effect due to the falling water. I am trying to make the point I think the $\dot{m}$ term in the equation $F= m\dot{v}+\dot{m}v=0$ has nothing to do with the $\dot{m}$ due to falling water and in this case of the cart it is null in the same way the $\dot{v}$ term is null.

 

[Delta2]

I am trying to make the point I think the m˙ term in the equation F=mv˙+m˙v=0 has nothing to do with the m˙ due to falling water

The $\dot m$ term is uniquely defined, I don't know how you say in one equation it is this, and in the other equation is that. Sorry you make no sense to me.

 

[bob012345]

The $\dot m$ term is uniquely defined, I don't know how you say in one equation it is this, and in the other equation is that. Sorry you make no sense to me.

I realize my point is subtle. The $\dot m$ term is uniquely defined. It is defined by the existence or not of a force. As I quoted Newton above, I think he meant that a force causes a momentum change not a momentum change causes a force. I think we all can agree there is no slowing down or speeding up, no force, on the cart due to water dripping. You then say there is a paradox if we look at the cart alone and can only resolve it by considering the falling water as well so everything works out to cancel any spurious forces. That makes no sense to me. I think it is more natural to say Newton's Second Law always works in every situation and on every part of a system but understanding and applying it is sometimes tricky. Of course, I could be all wet...

 

[Delta2]

That makes no sense to me. I think it is more natural to say Newton's Second Law always works in every situation and on every part of a system but understanding and applying it is sometimes tricky

And you make no sense to me sorry, you apply the law by defining $\dot m$ as zero, when it is not zero. Sorry that's way too tricky way to apply the law...