Damped oscillator with changing mass

[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!