Calculating Piston Oscillation Frequencies in a Gas-Filled Tube

[skrat]

Homework Statement

The inside of horizontal and $1m$ long tube is divided by 3 moving pistons (no friction) with $m=1kg$ into 4 identical parts - each containing $10g$ of Helium at constant temperature $T=300K$. Calculate the frequency of oscillation for each piston around the equilibrium position, if the only force responsible for any movements is due to the change of gas pressure.

Homework Equations

The Attempt at a Solution

Here is what I got and think is wrong:

$F=pS$ obviously.

$p=\frac{m}{MV}RT$


$F=\frac{m}{Mx}RT$

Now for each piston separately:
first one:

$-\frac{mRT}{M}(\frac{1}{x_1}+\frac{1}{x_2-x_1})=m\ddot{x_1}$

second:

$\frac{mRT}{M}(\frac{1}{x_2-x_1}-\frac{1}{x_3-x_2})=m\ddot{x_2}$

third:

$\frac{mRT}{M}(\frac{1}{x_3-x_2}-\frac{1}{x_3})=m\ddot{x_3}$

Where $x_{n}=x_{n0}+\varepsilon _n$.

Now even if this would be right, I have no idea how to continue.

 

[rude man]

You simultaneously have m = mass of the gas in one compartment AND the mass of one piston?
I assume M is the gram-molecular weight of He.

Net force on a piston is the difference in pressure on each side, times area. For small variations in piston position you can thus develop a simple 2nd order diff. eq. with constant coefficients to solve for x(t) where x is the distance a piston is displaced to one side of its equilibrium position.

EDIT: I forgot the other pistons!
Giive me a bit of time to set this up in my head ... & I'll try to analyze what you did also.

 

[skrat]

Yes, M is molecular wight of the He.
That's a typo. I will edit my first post and add index $0$ to the mass of the piston.

 

[skrat]

Now for each piston separately:
first one:

$-\frac{mRT}{M}(\frac{1}{x_1}+\frac{1}{x_2-x_1})=m_0\ddot{x_1}$

second:

$\frac{mRT}{M}(\frac{1}{x_2-x_1}-\frac{1}{x_3-x_2})=m_0\ddot{x_2}$

third:

$\frac{mRT}{M}(\frac{1}{x_3-x_2}-\frac{1}{x_3})=m_0\ddot{x_3}$

 

[rude man]

$F=\frac{m}{Mx}RT$

This says that F = infinity if x = 0 so that can't be right.

Linearize your 3 equations by using differentials: for isothermal process, pdV + Vdp = 0. From this develop 3 equations coupling the pistons' movements into each other. The equations are all linear 2nd order with constant coefficients but they have to be solved simultaneously.

 

[rude man]

The math looks horrible. I seem to have gotten 2 frequency components for the middle piston and 3 for the left and right one! Easy to make a math mistake here though. I think you should at least set up the three simultaneous ODE's.

 

[skrat]

This says that F = infinity if x = 0 so that can't be right.

Why would that be wrong?$x$ is measured from the very beginning of the tube. So if the displacement of the first piston is $\delta x$, than the total distance from the beginning of the tube is $x=x_1+\delta x$ where $x_1$ is equilibrium position of the first piston.

In other words: If you are strong enough to push the first piston towards the very beginning (so $x->0$) than the force on the piston would be enormous (mathematically speaking: $F->\infty$ if $x->0$). That makes sense if you ask me. Or .. ?

 

[Orodruin]

I suggest writing the differential equations in terms of the displacement from the equilibrium position rather than the positions relative to the beginning of the tube. This will make it much simpler to linearize the system for small oscillations.

 

[skrat]

Linearize your 3 equations by using differentials: for isothermal process, pdV + Vdp = 0. From this develop 3 equations coupling the pistons' movements into each other. The equations are all linear 2nd order with constant coefficients but they have to be solved simultaneously.

Hmm, I hope you meant this:

$Vdp+pdV=0$
$Vdp+pSdx=0$ I will use $ε=dx$ so$Vdp+pSε=0$
$dp=-\frac{pS}{V}ε$

$dF=dpS=-\frac{pS^2}{V}ε$

For the first piston:

$-\frac{pS^2}{V}ε_1+\frac{pS^2}{V}(ε_2-ε_1)=m_0\ddot{ε_1}$

 

[Orodruin]

Yes, something like that. Note the common prefactor. When you write down all of the equations, can you write them in terms of a matrix equation?

 

[BvU]

Two cents worth: Rudy signals horrifying math (and I am lazy and it's late here). So try something:
Cent 1:
What about a simpler problem, such as one piston only ?
Cent 2: I can imagine modes of oscillation where the outer two move in opposite phase and the middle one is stationary. Another mode has all three swinging in phase with the amplitudes of the outer ones the same and of the middle one something like a factor of 2 bigger.

My silent fear is there are several frequencies around...

 

[rude man]

Hmm, I hope you meant this:

$Vdp+pdV=0$
$Vdp+pSdx=0$ I will use $ε=dx$ so$Vdp+pSε=0$
$dp=-\frac{pS}{V}ε$

$dF=dpS=-\frac{pS^2}{V}ε$

For the first piston:

$-\frac{pS^2}{V}ε_1+\frac{pS^2}{V}(ε_2-ε_1)=m_0\ddot{ε_1}$

Exactly what I had in mind. In fact, what I did. So finish for pistons 2 and 3.
Can't you switch from epsilons to x's?

If you get the other two equations right I would give you 8/10 at least. Solving the 3 simultaneous ODE's is very messy. I think I got the frequencies but I did not complete the solutions for x1(t), x2(t) and x3(t). Some kind soul should have a shot at it, especially if they have wolfram alpha pro available.

 

[rude man]

My silent fear is there are several frequencies around...

I came up with 3, count 'em 3, frequencies.
The left and right pistons have 3 frequencies, the middle one two.

I'm too cheap to pay for pro Wolfram Alpha so if you have that facility why don't you carry out the math. I like using Laplace transforms but wolfram crapped out on me. The expressions are VERY messy, at least the ones I came up with.

 

[Orodruin]

Since there are three degrees of freedom in the problem there are going to be three frequencies. There is no need to use Laplace transforms to find them as it is a problem of the form
$$
\ddot x = - A x
$$
where x is a vector containing the degrees of freedom and A is a matrix.

OP: Try to write your system on this form, do you know how to solve it?

 

[rude man]

There is no need to use Laplace transforms to find them as it is a problem of the form

I said I liked using the Laplace transform. I didn't say one had to. The equations are ODE's , linear 2nd order, constant coefficients. Nice thing about the Laplace method is it changes the ODE's into algebraic equations, making it a simple system of 3 equations in 3 unknowns. À chacun son goût.

 

[skrat]

Exactly what I had in mind. In fact, what I did. So finish for pistons 2 and 3.
Can't you switch from epsilons to x's?

For you my friend I will! =)

If you get the other two equations right I would give you 8/10 at least. Solving the 3 simultaneous ODE's is very messy. I think I got the frequencies but I did not complete the solutions for x1(t), x2(t) and x3(t). Some kind soul should have a shot at it, especially if they have wolfram alpha pro available.

Can you please contact my assistant? :D

@Orodruin: Trying to write my system in matrix form and trying to solve it correctly:

no. 1: $-\frac{pS^2}{Vm}x_1+\frac{pS^2}{Vm}(x_2-x_1)=\ddot{x_1}$

$\ddot{x_1}+\frac{pS^2}{Vm}(2x_1-x_2)=0$

no. 2: $-\frac{pS^2}{Vm}(x_1)+\frac{pS^2}{Vm}(x_3-x_2)=\ddot{x_2}$

$\ddot{x_2}+\frac{pS^2}{Vm}(2x_2-x_1-x_3)=0$

no.3: $-\frac{pS^2}{Vm}(x_3-x_2)-\frac{pS^2}{Vm}x_3=\ddot{x_3}$

$\ddot{x_3}+\frac{pS^2}{Vm}(2x_3-x_2)=0$Using the matrix:

$\ddot{\vec{x}}+A\vec{x}=\ddot{\vec{x}}+\omega _0^2\tilde A \vec{x}=0$ where $\omega _0^2=\frac{pS^2}{Vm}$.

Looking for egienvalues and eigenvectors:

$det(\omega _0^2\tilde A-\omega ^2I)=det(\tilde A-\frac{\omega ^2}{\omega _0 ^2 }I)=0$

$det(\begin{bmatrix} 2-\lambda &-1 &0 \\ -1 & 2-\lambda & -1\\ 0& -1 &2-\lambda \end{bmatrix})=0$ leaves me with three eigenvalues and three eigenvectors:

a) $\lambda =\frac{\omega ^2}{\omega _0 ^2 }=2$ meaning $\omega =\sqrt 2 \omega _0$ with an eigenvector $\nu _1=(-1,0,1)$.

b)$\lambda =\frac{\omega ^2}{\omega _0 ^2 }=2+\sqrt 2$ meaning $\omega =\sqrt{2+\sqrt 2} \omega _0$ with an eigenvector $\nu _2=(1,-\sqrt 2,1)$.

c)$\lambda =\frac{\omega ^2}{\omega _0 ^2 }=2-\sqrt 2$ meaning $\omega =\sqrt{2-\sqrt 2} \omega _0$ with an eigenvector $\nu _2=(1,\sqrt 2,1)$.So If I am not mistaken these are all the frequencies of possible movements (described by eigenvectors) of this system.

Please correct me if wrong.

 

[Orodruin]

Looks good to me. I did not check the math in detail, but the end result looks very much like I would expect.

 

[rude man]

For you my friend I will! =)
Can you please contact my assistant? :Dno. 1: $-\frac{pS^2}{Vm}x_1+\frac{pS^2}{Vm}(x_2-x_1)=\ddot{x_1}$

$\ddot{x_1}+\frac{pS^2}{Vm}(2x_1-x_2)=0$

no. 2: $-\frac{pS^2}{Vm}(x_1)+\frac{pS^2}{Vm}(x_3-x_2)=\ddot{x_2}$

$\ddot{x_2}+\frac{pS^2}{Vm}(2x_2-x_1-x_3)=0$

no.3: $-\frac{pS^2}{Vm}(x_3-x_2)-\frac{pS^2}{Vm}x_3=\ddot{x_3}$

$\ddot{x_3}+\frac{pS^2}{Vm}(2x_3-x_2)=0$

Your 3 equations are correct and so are the frequencies. Very nicely done. BTW I consider this problem beyond the scope of introductory physics, mainly due to the advanced nature of the math.

P.S. to you or other posters: with the method of eigenvectors and eigenvalues, can it be determined which frequency components attach to which pistons? It's been a looong time since I had this stuff.

I had determined that the first and third pistons contained all three frequencies whereas the middle one vibrated at your (b) and (c) frequencies only.

 

[skrat]

P.S. to you or other posters: with the method of eigenvectors and eigenvalues, can it be determined which frequency components attach to which pistons? It's been a looong time since I had this stuff.

I had determined that the first and third pistons contained all three frequencies whereas the middle one vibrated at your (b) and (c) frequencies only.

Everything matches:

a) $\lambda =\frac{\omega ^2}{\omega _0 ^2 }=2$ meaning $\omega =\sqrt 2 \omega _0$ with an eigenvector $\nu _1=(-1,0,1)$.

b)$\lambda =\frac{\omega ^2}{\omega _0 ^2 }=2+\sqrt 2$ meaning $\omega =\sqrt{2+\sqrt 2} \omega _0$ with an eigenvector $\nu _2=(1,-\sqrt 2,1)$.

c)$\lambda =\frac{\omega ^2}{\omega _0 ^2 }=2-\sqrt 2$ meaning $\omega =\sqrt{2-\sqrt 2} \omega _0$ with an eigenvector $\nu _2=(1,\sqrt 2,1)$.First component of each eigenvector represents offset of first piston, second one of the second piston and the third component for the third piston.

Now if you look at the first components only, non of them is zero, meaning the first piston will always move and all three frequencies are possible.
The second components are a bit more tricky. In a) the second component is 0, meaning the second piston will never have the frequency that belongs to a). The other two frequencies are of course possible.

At least, that's what they learned me. There might be a more mathematical answer to your question, this is just how I understand it.Anyway, big thanks to all of you helping me!

 

[rude man]

That is good and interesting. Thanks.