I Damped oscillator with changing mass

[bob012345]

Where do you encounter the first issue that needs clarification? Between the two posts ( and accompanying diagrams ) I honestly thought I covered everything.

Are the $V_j, V_b$ terms independent of the oscillation?

 

[erobz]

Are the $V_j, V_b$ terms independent of the oscillation?

They are supposed to be, but I don't know...maybe that is a poor model. They are the velocity of water in the jet, and the water in the bucket relative to the bucket.
EDIT:
The way I thought about it was what comes out the hole was a function of the pressure at the bottom of the bucket. That pressure was dependent on the acceleration of the bucket. Those velocities are indirectly dependent on the bucket acceleration, through the pressure at the bottom and Bernoulli's principle.

 

[bob012345]

They are supposed to be, but I don't know...maybe that is a poor model. They are the velocity of water in the jet, and the water in the bucket relative to the bucket.
EDIT:
The way I thought about it was what comes out the hole was a function of the pressure at the bottom of the bucket. That pressure was dependent on the acceleration of the bucket. Those velocities are indirectly dependent on the bucket acceleration, through the pressure at the bottom and Bernoulli's principle.

I still recommend solving the water flowing from the still bucket first. One thing also, water is virtually incompressible so I wonder about assuming the pressure changes with acceleration? But obviously something changes and the rate water comes out must vary.

Here is a past PF thread dealing with that question;

https://www.physicsforums.com/threads/basic-calc-problem-with-flow-out-of-a-hole-in-a-bucket.840469/

 

[erobz]

One thing also, water is virtually incompressible so I wonder about assuming the pressure changes with acceleration?

I think it does. Imagine if you were sitting at the bottom of a tank of water of depth $z$. you are going to feel pressure $P = \rho g z$. Now, put that tank on a rocket ship. You are now going to feel pressure increase proportional to the rocket acceleration.

 

[bob012345]

One thing, the water leaking from the bucket should be less than normal maybe even zero when the bucket is accelerating downward. Does you equation physically allow for that?

 

[erobz]

One thing, the water leaking from the bucket should be less than normal maybe even zero when the bucket is accelerating downward. Does you equation physically allow for that?

I got this when I derived it (there is a possibility I got the sign convention mixed up), but yeah.

$$ P = - \rho \left( \frac{d^2x}{dt^2} - g\right) z $$

That result is substituted into this:

$$ \frac{dM}{dt} = - \rho A_j \sqrt{ \frac{2 P}{ \rho \left( 1 - \tau^2 \right) }} $$

So if $\frac{d^2x}{dt^2} \rightarrow g$, then $P \to 0$ and $\frac{d M(t)}{dt} \to 0$. The bucket stops leaking.

 

[erobz]

Yeah, I think that part is ok.

If we let $\ddot x = 0$ we recover $P = \rho g z$ ( static bucket )

If we let $\ddot x = g$ , we get P = 0 ( the bucket is in freefall )

and if we let the acceleration be negative $\ddot x < 0$ ( bucket accelerating upwards in my convention ), we get $P > \rho g z$

It seems reasonable to me.

 

[bob012345]

How do you distinguish between the changing water mass which can go to zero and the fixed bucket mass?

 

[erobz]

How do you distinguish between the changing water mass which can go to zero and the fixed bucket mass?

In the variable $M$ which is the sum of the two

$$ M(t) = m_b + m_w(t) $$

$$ \frac{dM(t)}{dt} = \frac{ d m_w(t)}{dt} $$

 

[bob012345]

In the variable $M$ which is the sum of the two

$$ M(t) = m_b + m_w(t) $$

I look forward to your numerical solution unless you wish to leave that to others.

 

[erobz]

I look forward to your numerical solution unless you wish to leave that to others.

Stil haven't considered the thrust force from the fluid jet. That changes the equations dramatically...the question is does it change the outcome?

I just wanted to see if I could properly set it up. I'm not sure how to continue. Linearization? It probably well beyond my technical know-how. How would you go about it?

 

[bob012345]

I just wanted to see if I could properly set it up. I'm not sure how to continue. Linearization? It probably well beyond my technical know-how. How would you go about it?

As far as the set up, I do know of a source you can compare these to but I didn't want to send it to you unless you really wanted to see it and compare notes. I don't want to spoil your day again :)

As for solving it, of course first try and simplify it as much as possible and then pick values for constants. Maybe make limiting case assumptions with the equations before trying to solve them as you derived them and solve those reduced equations first. For example, water leakage from a fixed bucket follows a parabola in time. If it is slow it might be approximated as linear. That simplifies the equations. Given a total mass and spring constant you know the oscillation frequency and can then approximate the mass flow rate variation with acceleration also. This is why I was suggesting having the limiting cases already solved first so one can make approximations.

As for what I would do to actually try and solve this beast? I would see if it can be modeled as a coupled electrical circuit or as an analog circuit and model it in a SPICE simulator like LTSpice since complex nonlinear circuits are fairly easy to get results in those kinds of simulators.

But I would ask for more expertise here at PF. Perhaps start a thread dedicated to solving this problem.

 

[bob012345]

This is just a paper I already pointed out yesterday but a better version (with color). I noticed it simplifies the physics to a more manageable form and uses a spreadsheet to numerically solve the equation. You might begin here and see how your equations align with or are different from this work.

https://www.researchgate.net/publication/264123215_The_motion_of_a_leaking_oscillator_A_study_for_the_physics_class

 

[erobz]

This is just a paper I already pointed out yesterday but a better version (with color). I noticed it simplifies the physics to a more manageable form and uses a spreadsheet to numerically solve the equation. You might begin here and see how your equations align with or are different from this work.

https://www.researchgate.net/publication/264123215_The_motion_of_a_leaking_oscillator_A_study_for_the_physics_class

Yeah, the only thing that "spoiled my day" getting the first solution was not being able to feel "justified" (my wife wouldn't call it justification) for slacking off on the housework, which is why I felt compelled to go forward from there today!

 

[erobz]

But I would ask for more expertise here at PF. Perhaps start a thread dedicated to solving this problem.

Well, I'll wait it out. Technically, this is still a simplification to the OP's inquiry which involved air drag. Also, several others hinted at the form of the driving equations already. Hopefully, someone will confirm or deny if this is indeed what they had in mind.

 

[bob012345]

Well, I'll wait it out. Technically, this is still a simplification to the OP's inquiry which involved air drag. Also, several others hinted at the form of the driving equations already. Hopefully, someone will confirm or deny if this is indeed what they had in mind.

I guess it all depends on how one sets up the experiment. A smallish plastic cup with a light spring and a small hole might have air resistance a bigger factor than the dripping water whereas a bucket with a gallon of water moving slowly might have little air resistance.

I'm not sure what you mean by driving equations.

 

[erobz]

Just so there is less concealed, this is how I derived the pressure relationship:

I applied Newtons 2nd for Accelerating frame:

$$ \sum F - m \ddot x = m \ddot z $$

Where;

$\ddot x$ is the acceleration of the oscillator
$\ddot z$ is the acceleration of the fluid element in the accelerating frame

From here I chose to make some simplifying assumption that the area of the cup is much larger than the jet so that $\ddot z$ can be neglected.

This implies that:

$$ \sum F = m \ddot x $$

$$ PA - ( P + dP )A + \rho A g dz = \rho A dz \frac{d^2 x}{dt^2} $$

This simplifies to:

$$ \frac{dP}{dz} = - \rho \left( \frac{d^2 x}{dt^2} - g \right) $$

Integrating that from the free surface of the water to the bottom of the cup gives the result:

$$ P = - \rho \left( \frac{d^2 x}{dt^2} - g \right) z $$
___________________________________________________________________________________________________________________

If you don't make this simplifying assumption above, you add a third ODE to the system, and it becomes even more hideous:

$$ \frac{dP}{dz} = - \rho \left( \frac{d^2 x}{dt^2} + \frac{d^2 z}{dt^2} - g \right) $$

Remembering:

$$ z = \frac{M(t) - m_b}{A_b} $$

It follows that:

$$ \frac{d^2 z}{dt^2} = \frac{1}{A_b} \frac{d^2 }{dt^2} \left( M(t) \right) $$

and

$$ \frac{d}{dz} \left( M(t) \right) = A_b $$

I find that the resulting equation is given by:

$$ A_b \frac{dP}{dt} = - \rho \left( \frac{d^2 x}{dt^2} + \frac{1}{A_b} \frac{d^2 }{dt^2} \left( M(t) \right) - g \right) \frac{d}{dt} ( M(t) ) $$

 

[erobz]

I guess it all depends on how one sets up the experiment. A smallish plastic cup with a light spring and a small hole might have air resistance a bigger factor than the dripping water whereas a bucket with a gallon of water moving slowly might have little air resistance.

I'm not sure what you mean by driving equations.

Maybe "governing equations" was the proper terminology.

 

[bob012345]

Could you please clarify one thing. I am a bit confused over the difference between $x$ and $x_0(t)$.

$$M(t) \frac{d^2}{dt^2} \left( x- x_o(t) \right) = M(t) g - k \left( x- x_o(t) \right) \tag{1}$$

 

[erobz]

Could you please clarify one thing. I am a bit confused over the difference between $x$ and $x_0(t)$.

$$M(t) \frac{d^2}{dt^2} \left( x- x_o(t) \right) = M(t) g - k \left( x- x_o(t) \right) \tag{1}$$

$x_o(t)$ is the equilibrium position of the oscillator as a function of time. In normal oscillator problems $x_o$ is static, but here it is time dependent ( and actually is just a function of $M(t)$). The difference $x - x_o$ is the displacement of the system from equilibrium. I think.

 

[bob012345]

$x_o(t)$ is the equilibrium position of the oscillator as a function of time. In normal oscillator problems $x_o$ is static, but here it is time dependent ( and actually is just a function of $M(t)$). The difference $x - x_o$ is the displacement of the system from equilibrium. I think.

I see. Simply the level it oscillates around changes as the system ejects mass. I suppose it gets higher in this problem.

 

[erobz]

Cleaning them up, and dropping the function notation:

$$ M \ddot x + k x = \left( 2 + \frac{\ddot M}{k} \right) g M \tag{1} $$

$$ \ddot x = \beta \frac{ {\dot M}^2}{M - m_b} + g \tag{2} $$

Where $\beta$ is a constant given by:

$$ \beta = - \frac{1}{ 2 \rho A_j}\frac{ \left( 1 - \tau^2 \right) }{ \tau } $$

and $\tau$ is the ratio of the cross sectional area of the jet to the bucket:

$$ \tau = \frac{A_j}{A_b}$$

 

[erobz]

Could you please clarify one thing. I am a bit confused over the difference between $x$ and $x_0(t)$.

$$M(t) \frac{d^2}{dt^2} \left( x- x_o(t) \right) = M(t) g - k \left( x- x_o(t) \right) \tag{1}$$

I think I found an error in (1). That $2$ in the factor on the RHS set off some alarm bells after I tried to think about the initial conditions.

The way I have the coordinate system (1) should actually be...( I think ):

$$M(t) \frac{d^2}{dt^2} \left( x- x_o(t) \right) = M(t) g - k \left( x_o(t) + \left[ x - x_o(t) \right] \right) \tag{1}$$

Which means the tidied up equation is:

$$ M \ddot x + k x = \left( 1 + \frac{\ddot M}{k} \right) M g \tag{1} $$

Never trust me Bob! Sorry.

 

[bob012345]

I think I found an error in (1). That $2$ in the factor on the RHS set off some alarm bells after I tried to think about the initial conditions.

The way I have the coordinate system (1) should actually be...( I think ):

$$M(t) \frac{d^2}{dt^2} \left( x- x_o(t) \right) = M(t) g - k \left( x_o(t) + \left[ x - x_o(t) \right] \right) \tag{1}$$

Which means the tidied up equation is:

$$ M \ddot x + k x = \left( 1 + \frac{\ddot M}{k} \right) M g \tag{1} $$

Never trust me Bob! Sorry.

If you can bear it, I think that equation is quite close to one of the two equations (eq, 23) in this princeton treatment;

https://www.hep.princeton.edu/~mcdonald/examples/bucket_osc.pdf

 

[erobz]

If you can bear it, I think that equation is quite close to one of the two equations (eq, 23) in this princeton treatment;

https://www.hep.princeton.edu/~mcdonald/examples/bucket_osc.pdf

Yeah, this link seems to be more rigorous.

I've got to be honest, I still don't think (1) is correct. I'm losing it!

 

[bob012345]

Yeah, this link seems to be more rigorous.

I've got to be honest, I still don't think (1) is correct. I'm losing it!

No need to lose it! It's actually kind of a complicated problem. You have a few other treatments to compare to and I think they are not all exactly the same which is to be expected.

 

[erobz]

It's hard to do a one-to-one comparison. For starters they assume a spring a zero free length for "simplicity" ( I find that somewhat amusing given the analysis that follows). I understand their motivation for that, they only care how it moves, not where it moves. Anyhow, seeing that assumption helped me to question what I was messing up in (1). ( the free length of the spring )

Let $l_o$ be the free length of the spring, and $s(t)$ be the required deflection for equilibrium:

$$ x_o(t) = l_o + s(t) $$

From this (1) becomes:

$$ \begin{align} M \frac{d^2}{dt^2} \left( x - x_o(t) \right) &= -k \left( s + ( x - ( l_o + s ) ) \right) + Mg \tag*{} \\ M \frac{d^2}{dt^2} \left( x - ( l_o + s) \right) &= -k \left( x - l_o \right) + Mg \tag*{} \\ M \frac{d^2}{dt^2} \left( x - s \right) &= -k \left( x - l_o \right) + Mg \tag*{} \end{align} $$

With

$$ s = \frac{Mg}{k} $$

It follows that (1) is given by:

$$ M \ddot x + k x = \left( 1 + \frac{ \ddot M}{k} \right) M g + k l_o \tag{1} $$

As a verification if we plug in that the hole is plugged $\dot M = 0$ and $x = x_o$ we find that:

$$ M \ddot x + k \left( l_o + s\right) = Mg + k l_o $$

$$ M \ddot x = Mg - ks $$

but $Mg - ks \equiv 0$

So as we expect, the system remains static, i.e. $M \ddot x = 0$

I'm more settled on that now.

Also, the authors in the Princeton paper are including the "thrust force" from the jet.

Furthermore, they do a correction for Bernoulli's in an accelerated frame, but it's a little complex for me to tell what is happening there in comparison to how I "tried" to correct for it.

 

[vanhees71]

The treatment in the Princeton paper looks correct to me.

 

[erobz]

The treatment in the Princeton paper looks correct to me.

I can't be "sure" they are correct, because I haven't fully comprehended it, but I'll take your (and their) word on that. My problem is seeing where I am not correct based on their work. Thats where I was hoping to have a "live" dialog here with someone who understands it ( can translate it ).

 

[erobz]

This doesn't relate to what's wrong with my problem ( as it is the consideration of the fluid jet)

in [1] the authors derive a $F_{react,z}$ (18) & (19) and it seems like its incorrect.

They have that:$$ F_{react,z} = \rho a h \left ( \dot z - \frac{A}{a} \dot h \right) \tag{19}$$

The units don't match.

The use $V$ for the velocity of the jet, $a$ for the area of the hole, and $A$ for the area of the bucket.I think it should instead be:$$ F_{react,z} = \rho A \dot h \left ( \dot z - \frac{A}{a} \dot h \right) $$

Maybe its just a typo?
[1] https://www.hep.princeton.edu/~mcdonald/examples/bucket_osc.pdf

 

[erobz]

Yeah, it's a typo:

$$ \frac{dp}{dt} = F_{ext} + F_{reation} $$

They have that the total momentum of the system $p$ (considered as the bucket and the water in the bucket)

$$ p = m_b \dot z + \rho A h \left( \dot z - \dot h \right) $$

$$\frac{dp}{dt} = m_b \ddot z + \rho A \dot h \dot z + \rho A h \ddot z - \rho A \dot h^2 - \rho A h \ddot h $$

$$ F_{ext} + F_{reation} = \left( m_b + \rho A h \right) g - kz + \rho A \dot h \left ( \dot z - \frac{A}{a} \dot h \right) $$$$ m_b \ddot z + \rho A \dot h \dot z + \rho A h \ddot z - \rho A \dot h^2 - \rho A h \ddot h = \left( m_b + \rho A h \right) g - kz + \rho A \dot h \left ( \dot z - \frac{A}{a} \dot h \right) $$

After all that is simplified I arrive at the authors ( 21 ).

 

[erobz]

Well, I think I understand most of the relevant bits of the material in the Princeton paper. Their solution for when $a << A$ is basically the result I believe I would get if I ignored the pressure correction of Bernoulli's I tried to make in the equation for the mass flowrate out of the bucket (they both ignore the jet reaction force for that case). They don't try to propose a solution form for the genera case when $0 < \frac{a}{A} < 1$, but finding a solution to that is over my head regardless.I hope the OP got something out of my struggle to understand.

Thank you all for your input, especially @bob012345 for tracking down these solutions!