I Damped oscillator with changing mass

[Delta2]

Is (1) the proper differential equation?

In my opinion it is proper only if we consider that the water exits the bucket with negligible relative velocity, i.e just dripping. Because then (1) will be what we get from the rocket equation.

 

[erobz]

In my opinion it is proper only if we consider that the water exits the bucket with negligible relative velocity, i.e just dripping. Because then (1) will be what we get from the rocket equation.

Agreed, but I just wanted to make sure I set the simplest version up correctly, before trying to add the complexity of the fluid jet which will have implications in the thrust force, and the rate of change of mass. Also, if we fully consider it, I think we have to account for how the acceleration of the oscillator impacts the jet and also then the mass flow rate out of the bucket. I think that is where the "coupled" differential equations come in that @Orodruin mentioned.

 

[Delta2]

BTW with what software did you make the figure of post #90, looks pretty decent!

 

[erobz]

BTW with what software did you make the figure of post #90, looks pretty decent!

PowerPoint. Its gotten a little more diagram friendly over the years with the addition of some "object snapping"

 

[pnachtwey]

OK, you have the acceleration of the bucket. You still need another differential equation for how the water flows out the orifice on the bottom. The water will flow out faster when the bucket is accelerating upwards and flow out slower when the bucket is decelerating. You still need two differential equations. One for the position of the bucket, which you have and another for the change in mass of the bucket do to water flowing out

 

[erobz]

OK, you have the acceleration of the bucket. You still need another differential equation for how the water flows out the orifice on the bottom. The water will flow out faster when the bucket is accelerating upwards and flow out slower when the bucket is decelerating. You still need two differential equations. One for the position of the bucket, which you have and another for the change in mass of the bucket do to water flowing out

I'm planning on working my way up to that. I had to mow the grass first!

I'm taking it you think the effect of the acceleration of the oscillator on the mass flow rate through the hole is larger than the effect of the thrust from the fluid jet?

 

[bob012345]

You could try letting the mass oscillate side to side yet dripping from the bottom first as an easier case.

 

[erobz]

Focusing in on the draining water. $z$ is measured down from the free surface ( assumed to be at $P_{atm}$ ). Neglecting viscous losses, apply Bernoulli's across the hole:

$$ \frac{P}{\rho g} + \frac{V_b^2}{2g} + z_b = \frac{P_{atm}}{\rho g} + \frac{V_j^2}{2g} + z_h $$

$z_b \approx z_h$

$A_j V_j = A_b V_b \implies V_b = \frac{A_j}{A_b} V_j = \tau V_j$

We are left with:

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

Furthermore, the rate of change in oscillator mass $M$ is the rate of change of the liquid mass

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

In the accelerated frame of reference of the bucket I get the following result for the pressure at the base of the bucket:

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

And using the total mass of the system $M$ to eliminate $z$ we have that:

$$ z = \frac{M - m_b}{ \rho A_b} $$

So the final result for the mass flow rate is given by:

$$ \frac{dM(t)}{dt} = - A_j \sqrt{ \frac{ - 2 \rho \left( \frac{d^2x}{dt^2} - g\right) }{ \left( 1 - \tau^2 \right) } \frac{M(t) - m_b}{ A_b} } \tag{2} $$

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

If I have it correct, (1) and (2) are a nasty pair of coupled non-linear second order ODE's?

 

[bob012345]

It's not clear what all of your notation is. Could you please describe what's what? Thanks.

Also, you might solve for the dripping water first with a fixed bucket to see the form of that.

 

[erobz]

It's not clear what all of your notation is. Could you please describe what's what? Thanks.

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

Also, you might solve for the dripping water first with a fixed bucket to see the form of that.

What problem do you want me to solve? A stationary bucket dripping water?

 

[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.