Damped oscillator with changing mass

[Rezex124]

TL;DR Summary

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

Hello,

So about two weeks ago in class we looked at RLC circuits in our E&M course, and short story short... we compared the exchange of energy between the Capacitor and the Inductor (both ideal) to simple harmonic motion. Once the capacitor and inductor are not ideal anymore, we said it's analogue to damped harmonic oscillations.

I understood what we were doing and I could work with the equations, but we have yet to have a course on oscillations and waves, so I didn't really know where the equations came from. So naturally I went and researched it a bit in my own time. I found out what I was looking for, and to my surprise - I really liked learning about oscillations !
Over the last few days, I did some experiments in our school lab as well, to compare theory and real world. I was playing around with all sorts of combinations, changing the spring or the mass etc. Then it got me thinking. For these cases I was only using weights with constant mass, and I asked myself how the experiment would look like, if the mass were changing.I came up with an experiment: I took a normal (yoghurt) plastic cup and made a little hole at the bottom. Then I hooked it up to a spring, covered the hole and poured water inside. I displaced it from its equilibrium position, uncovered the hole and simultaneously started my timer. Now I had a damped oscillating motion with changing mass !

After a few measurements I wanted to compare it to theoretical results, but I couldn't find anywhere on how to get the oscillation time. So I tried to derive it myself. But seeing as I am a first year undergrad student, I found out this is a bit out of my scope of knowledge.I tried this:

Net force is equal to change in momentum (ΣF=dpdt, where p = m*v). Now we all know if the mass is constant, we get Newton's 2nd law in it's most known form, but in my case, mass also changes with time. So: dpdt=ΣF=m∗dvdt+v∗dmdt.
Then I look at what forces act on my system (the cup):

• Force of gravity... fg=m(t)*g
• Air resistance... fu=-c*v*|v|, where c is a constant, determined experimentally
• Force because of the spring... fv=-kx

When I had all the forces, I wanted to see how mass changes with time. I defined mass (m) as: m=M+mw, where M is mass of the cup and mw mass of water remaining. Mass is volume times the density of material, so mw=ρ∗V (volume in this case is a function of time). I wrote out V = S*h, where S is the current area of water (at top) and h the current height of water left (reference point at bottom of the cup).
S=S0+S1−S0H∗h, where S0 is area of the hole made, S1 The starting area and H the starting height. Putting it all together, we get:
m=M+ρ∗hH∗[S0∗(H−h)+S1∗h]. and
dmdt=ρ∗[S0+(S1−S0)∗(2hH)]∗dhdt

I also got the velocity, at which the water was flowing out, how the volume changes over time and some others.

Now this was my starting point. I have some equations about my system, but I couldn't figure out how to combine them and get what I was looking for - the oscillation time.
This I where I got stopped in my tracks and I've been going in circles for a few days looking for something new.
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

 

[Delta2]

If you can determine (by making some simplifying assumptions, because I think the way the mass changes is coupled to the oscillatory motion but I might be wrong) the function $m(t)$ , then the ODE governing your system is $$-kx-m(t)g-c(\frac{dx}{dt})^2=m(t)\frac{d^2x}{dt^2}+\frac{dm}{dt}\frac{dx}{dt}$$. This is a non linear ODE with non constant coefficients, which in simple words means it is hard to solve, but even if you make it linear by taking the air resistance term to be $-c\frac{dx}{dt}$, you still might need numerical methods to solve it, depending on the form (simple or complex) of $m(t)$.

EDIT: NOT SURE if you should add the term $\frac{dm}{dt}\frac{dx}{dt}$ at the RHS of this ODE, to be honest I am not good in applications of Newton's 2nd law where mass is variable.

 

[Rezex124]

Thanks for the reply,

and yes I'm not so sure about that term either (not so experienced as well :P). But seeing both mass and velocity are functions of time, and you differentiate over time, I would expect that the (dm/dt) * (dx/dt) is also present. Plus it makes sense you have a term that tells you how quickly/slowly the mass changes.
So I'll leave it in for now, but just to be sure I will research it some more.

Also we haven't had a course in computational physics (all these classes are next semester), so I'm not really familiar on how to solve problems numerically. Do you have any suggestions in what program / programming language I should try to solve this?

 

[Delta2]

There is a variety of software for numerical solutions to differential equations. Undergraduate students (but not only) usually use MatLab or Maple , while professionals use COMSOL platform or other professional software. Take a look at the subforum here
https://www.physicsforums.com/forums/matlab-maple-mathematica-latex.189/

 

[Rezex124]

Great ! Thank you very much

 

[Delta2]

Paging @haruspex, he will probably have some additional valuable comment for your work.

 

[haruspex]

If you can determine (by making some simplifying assumptions, because I think the way the mass changes is coupled to the oscillatory motion but I might be wrong) the function $m(t)$ , then the ODE governing your system is $$-kx-m(t)g-c(\frac{dx}{dt})^2=m(t)\frac{d^2x}{dt^2}+\frac{dm}{dt}\frac{dx}{dt}$$. This is a non linear ODE with non constant coefficients, which in simple words means it is hard to solve, but even if you make it linear by taking the air resistance term to be $-c\frac{dx}{dt}$, you still might need numerical methods to solve it, depending on the form (simple or complex) of $m(t)$.

EDIT: NOT SURE if you should add the term $\frac{dm}{dt}\frac{dx}{dt}$ at the RHS of this ODE, to be honest I am not good in applications of Newton's 2nd law where mass is variable.

If c is the damping coefficient, I would make the dependence on speed linear, not quadratic. The speeds should be quite modest. (If quadratic it's messy; the sign would have to keep changing to oppose motion.)

I don't understand the $\dot m\dot x$ term.

For m(t), I think we can ignore the effect of the oscillation and just treat it as a static draining tank, $\dot h=-\rho \sqrt {gh}$. The accelerations will be much smaller than g.

 

[Delta2]

If c is the damping coefficient, I would make the dependence on speed linear, not quadratic. The speeds should be quite modest. (If quadratic it's messy; the sign would have to keep changing to oppose motion.)

hm yes I should 've written the quadratic term as $$-c\frac{dx}{dt}\left |\frac{dx}{dt}\right |$$.

I don't understand the $\dot m\dot x$ term.

Me neither e hehe but how to express that momentum is leaving the cup because mass is leaving.

For m(t), I think we can ignore the effect of the oscillation and just treat it as a static draining tank, $\dot h=-\rho \sqrt {gh}$. The accelerations will be much smaller than g.

I see thanks, I think even for this $m(t)$ and if we make the ODE linear it still doesn't have closed form solution.

 

[haruspex]

how to express that momentum is leaving the cup because mass is leaving.

The water leaving takes its share of momentum with it. That exerts no force on the cup.

 

[alan123hk]

I'm also not sure if $\frac{dm}{dt}\frac{dx}{dt}$ should exist.

On the other hand, note that $m(t)$ can be a relatively complex nonlinear equation, which may involve multiple factors, such as the amount of water remaining in the cup, the velocity of the cup, and the diameter of the bottom hole of the cup. This is what fluid mechanics deals with.

This is really an interesting experiment. It would be even more gratifying and satisfying if the accuracy of the equation could be verified experimentally.

 

[Rezex124]

For m(t), I think we can ignore the effect of the oscillation and just treat it as a static draining tank, $\dot h=-\rho \sqrt {gh}$. The accelerations will be much smaller than g.

I tried to duplicate the experiment (quickly) at my home - so I would pay more attention to the water flow. It wasn't an accurate experiment, but yes, I believe treating the leaving water as static drain should be fine.

However I don't understand where you got $\dot h = - \rho \sqrt{gh}$ from.
I got that $\dot V = - \phi _v = -S*v_w = -S\dot h$, which then follows that $\dot h = -v_w$. Then using Bernoulli's equation, I get that $\dot h = -\sqrt{2gh}$.

On the other hand, note that $m(t)$ can be a relatively complex nonlinear equation, which may involve multiple factors, such as the amount of water remaining in the cup, the velocity of the cup, and the diameter of the bottom hole of the cup. This is what fluid mechanics deals with.

Yes indeed, $m(t) = M + \rho \left[ S0 + \frac{S1-S0}{H}*2h(t) \right] h(t)$, amount of water left in cup is hidden in the h(t) term, which I can't (yet) seem to figure out how to express.

I don't understand the $\dot m\dot x$ term.

I will try to reach my professor about the $\dot m \dot x$ term. Maybe he has some ideas.Also thanks everyone for taking the time, you are a great help :)

 

[haruspex]

where you got h˙=−ρgh from.

The $\rho$ was just an arbitrary constant, so this agrees with your result.

I will try to reach my professor about the m˙x˙ term

Um, why? It was Delta2 that introduced it, and he agrees he did not have a sound reason for it.

 

[Rezex124]

Um, why? It was Delta2 that introduced it, and he agrees he did not have a sound reason for it.

Actually, I already had that term included when I was first asking the question (also wrote it in), but then most of you said it shouldn't be in, although I am still conflicted about it. On one side it tells you the rate of change of mass, but on the other change in mass is already considered in the $m(t)$.
Still, getting more opinions on this wouldn't hurt.

 

[haruspex]

Actually, I already had that term included when I was first asking the question (also wrote it in), but then most of you said it shouldn't be in, although I am still conflicted about it. On one side it tells you the rate of change of mass, but on the other change in mass is already considered in the $m(t)$.
Still, getting more opinions on this wouldn't hurt.

I do see $F=m\dot v+v\dot m$, which I strongly discourage. It is responsible for a great deal of confusion and serves no useful purpose. It would be true of a closed system, but in the real world mass is conserved in closed systems.

Consider e.g. a leaking water cart moving along a rail line at speed v. There appears to be no external horizontal force on it and $\dot m<0$, so from that equation it would appear that $\dot v>0$.

 

[Orodruin]

There are several complicating factors here that imply you cannot just go around changing the mass. In classical mechanics, mass is conserved and so mass needs to leave your system in order for it to decrease. When mass leaves the system, it will generally take some amount of momentum with it. You therefore cannot write down an equation governing your experiment unless you also model how the mass leaves the system. If the mass leaves at velocity $v$, it will take with it a momentum $-\dot m v$ with it. In order to model this properly you need to model both $m$ as a function of time (which could also be dependent on the motion itself, possibly with time delay) as well as the velocity $v$ (which will generally also depend on the motion).

Long story short, it will likely be quite a mess for anything but the most basic assumptions. The question is if these assumptions are sufficient to give a reasonable description of the system.

 

[sophiecentaur]

Just imagine that the water was dumped in packets, just when the cup reaches the bottom of its motion. No loss of Kinetic Energy at each dumping and it would make the calculation much easier. This is a possibly ideal but unlikely model but it would give you individual cycles of SHM with period
T_n = 2π√(m_n/k)
The 'rate of loss of m' would not be uniform, of course, but the error would be less as the rate of loss of m decreases. It would give a ball park result with which to compare the more formal result.

 

[haruspex]

Just imagine that the water was dumped in packets, just when the cup reaches the bottom of its motion. No loss of Kinetic Energy at each dumping and it would make the calculation much easier. This is a possibly ideal but unlikely model but it would give you individual cycles of SHM with period
T_n = 2π√(m_n/k)
The 'rate of loss of m' would not be uniform, of course, but the error would be less as the rate of loss of m decreases. It would give a ball park result with which to compare the more formal result.

Perhaps a bit better would be to take the standard solution for constant mass then plug in how the mass changes with time.

 

[Delta2]

Consider e.g. a leaking water cart moving along a rail line at speed v. There appears to be no external horizontal force on it and m˙<0, so from that equation it would appear that v˙>0.

I don't see anything wrong with the acceleration being non zero, the water leaving the cart act as some sort of exhaust (like in rockets) that gives additional velocity to the cart.

 

[haruspex]

I don't see anything wrong with the acceleration being non zero, the water leaving the cart act as some sort of exhaust (like in rockets) that gives additional velocity to the cart.

It is merely leaking. Dripping down.

 

[Delta2]

Someone needs to write an insight for Newton's 2nd when the mass is variable...

 

[sophiecentaur]

Perhaps a bit better would be to take the standard solution for constant mass then plug in how the mass changes with time.

I don't see anything wrong with the acceleration being non zero, the water leaving the cart act as some sort of exhaust (like in rockets) that gives additional velocity to the cart.

Both of those approaches involve mass variation at work all the time. Even the rate of dripping will be proportional to the instantaneous acceleration (zero at the equilibrium height). My back of a fag packet approach provides a exact solution to a dodgy model. I have to say, I'm not proud of it; just clutching at a straw.

 

[Orodruin]

Consider e.g. a leaking water cart moving along a rail line at speed v. There appears to be no external horizontal force on it and $\dot m<0$, so from that equation it would appear that $\dot v>0$.

This is - as many times - only a problem if one misinterprets that equation. With $F$ being the rate of momentum change in the system, the case of just dripping does not satisfy $F=0$ because momentum is leaking out of the system with the mass at a rate of $-\dot m u$, where $u$ is the velocity of the leaking mass. For $u=v$ we obtain $\dot v=0$.

For $u=v+u_0$, we do end up with an acceleration of $-[d\ln (m/m_0)/dt] u_0$.

 

[Delta2]

Something tells me that the ODE in post #2 is not correct, with or without the $\dot m v$ term and we need to take into account the "exhaust" velocity of the water in order to form a correct ODE.

 

[alan123hk]

We can refer to how the rocket equation handles this.
https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Dourmashkin)/12:_Momentum_and_the_Flow_of_Mass/12.03:_Rocket_Propulsion

A seemingly similar equation also appears in the derivation of the rocket equation as follows.

But in this equation $V$ and $v_e$ are two different velocities, $V$ is the velocity of the rocket relative to the observer and $v_e$ is the relative velocity of the matter leaving the rocket,

And the problem we deal with in this thread is more complicated, because the external force of the system is not a simple constant gravitational field, but is related to a complex oscillating system, which may need to be described by nonlinear differential equations, that is the term $~ c {\left( \frac {dx} {dt} \right) }^2 ~$ in the equation.

 

[haruspex]

This is - as many times - only a problem if one misinterprets that equation. With $F$ being the rate of momentum change in the system, the case of just dripping does not satisfy $F=0$ because momentum is leaking out of the system with the mass at a rate of $-\dot m u$, where $u$ is the velocity of the leaking mass. For $u=v$ we obtain $\dot v=0$.

For $u=v+u_0$, we do end up with an acceleration of $-[d\ln (m/m_0)/dt] u_0$.

Right, but I have never seen a text that states that matter leaving the system and taking momentum with it has to be treated as an external force. What I have seen is many students get problems wrong by blindly using that equation.

 

[Delta2]

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.

I think if we remove the term $\frac{dm}{dt}\frac{dx}{dt}$ the ODE of post #2 holds as an approximation in the limit that the exhaust velocity of water is the same as the velocity of the rest of the cup+water.

 

[alan123hk]

Assuming the system oscillates up and down (I'm not sure if OP means side to side). Is it possible that this is the case, although I made other bold assumptions in deriving the equation?

 

[Delta2]

Assuming the system oscillates up and down (I'm not sure if OP means side to side). Is it possible that this is the case, although I made other bold assumptions in deriving the equation?

View attachment 302477​

That's interesting derivation. Looks like $m(t)$ is coupled to $v(t)$ which makes the ODE highly non linear not only due to the $(\frac{dx}{dt})^2$ but due to terms $m\frac{dv}{dt}$ and $v\frac{dm}{dt}$

 

[alan123hk]

the ODE highly non linear not only due to the (dxdt)2 but due to terms mdvdt and...

Yes, whether the derivation is reasonable, right or wrong, these equations are too cumbersome and complex to be almost certainly impossible to solve analytically, and thus should be completely unappealing to the vast majority of people.

In other words, solving these equations for such a complex system is really a numerical analysis and engineering problem, not a physical one.

 

[Delta2]

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

EDIT: I am not sure at all, wikipedia calls it the effective exhaust velocity...

 

[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