Pendulum: two masses hanged on a weightless line with a spring

[skrat]

Homework Statement

I have a system of two masses m hanged with lines l from the ceiling. We put a spring between the two masses. Find a general solution for this pendulum for small angles

Homework Equations

The Attempt at a Solution

Hmmm..

$\sum M=0$

$J\ddot{\varphi _1}+mglsin\varphi _1+kxcos\varphi _1=0$,

for small angles $sin\varphi =\varphi$ and $cos\varphi =1$ and $x=l\varphi$, therefore:
$J\ddot{\varphi _1}+mgl\varphi _1+kl\varphi _1=0$

$\ddot{\varphi _1 }+\frac{(mg+k)}{ml}\varphi _1=0$

Symmetrically for the other body: $\ddot{\varphi _2 }+\frac{(mg+k)}{ml}\varphi _2=0$

is everything ok so far? Now... Well, the instructions say that I should get a system of differential equations and find a general solution.. So, my question is... How are these two equations above together a system - How are they "connected".

 

[Kosomoko]

You should look again at the term in your equation for the force from the spring. For each mass, the force from the spring depends on both φ₁ and φ₂

Something else occurs to me, you need to know the equilibrium length of the spring and how it compares to the separation between the two masses. I am guessing that you're meant to suppose that the equilibrium length of the spring is equal to the spacing of the two masses if they were at rest. Does the question say anything about this?

 

[skrat]

aaaaaaa so the force of the spring is actually $kx=k(x_1-x_2)= kl(\varphi _1-\varphi _2)$ ?

 

[Kosomoko]

Yes... assuming that the equilibrium length of the spring is equal to the distance between the two masses when they're at rest.

 

[skrat]

Thanks! :)

 

[Kosomoko]

You are most welcome.

 

[skrat]

Ok, I tried to find a general solution of this differential system:

Firstly, I made one correction concerning the force of the spring $kx=k(x_1+x_2)$ that is if at the begging we pull the first mass for $x_1$ away and the second mass for $x_2$ away. Total expansion of the spring is therefore $x_1+x_2=l(\varphi _1 + \varphi _2$ where one of the angles is negative.

So for first body:

$-\ddot{\varphi _1}+\varphi _1(\frac{g}{l}+\frac{k}{m})+\varphi _2\frac{k}{m}=0$

and second:

$-\ddot{\varphi _2}+\varphi _2(\frac{g}{l}+\frac{k}{m})+\varphi _1\frac{k}{m}=0$

This gives me a system : $\begin{bmatrix} \ddot{\varphi _1}\\ \ddot{\varphi _2} \end{bmatrix}=\begin{bmatrix} \frac{g}{l}+\frac{k}{m} &\frac{k}{m}\\ \frac{g}{l}+\frac{k}{m} & \frac{k}{m} \end{bmatrix}\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}$

Let's say that $\frac{g}{l}+\frac{k}{m}=\alpha$ and $\frac{k}{m}=\beta$

than $\begin{bmatrix} \frac{g}{l}+\frac{k}{m} &\frac{k}{m}\\ \frac{g}{l}+\frac{k}{m} & \frac{k}{m} \end{bmatrix}=\begin{bmatrix} \alpha &\beta\\ \alpha & \beta \end{bmatrix}$

Egien values for this matrix: $(\alpha -\lambda )(\beta -\lambda )-\alpha \beta =\lambda (\lambda -(\alpha +\beta ))=0$

Diagonal matrix is therefore $\begin{bmatrix} 0 & 0\\ 0 & \alpha +\beta \end{bmatrix}$ and eigenvector for $\lambda =0$:

$\begin{bmatrix} \alpha &\beta\\ \alpha & \beta \end{bmatrix}\sim \begin{bmatrix} \alpha &\beta\\ 0& 0 \end{bmatrix}$ so $v_1=(-\frac{\beta }{\alpha },1)$

and for $\lambda =\alpha + \beta$:

$\begin{bmatrix} -\beta &\beta\\ \alpha & -\alpha \end{bmatrix}\sim \begin{bmatrix} -\beta \alpha &\beta \alpha\\ \alpha \beta& -\alpha\beta \end{bmatrix}\sim \begin{bmatrix} -\beta \alpha &\beta \alpha\\ 0& 0 \end{bmatrix}$ therefore $v_1=(1,1)$

and accordingly matrix $P=\begin{bmatrix} -\frac{\beta }{\alpha } &1 \\ 1 & 1 \end{bmatrix}$

So

$\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0 & \alpha +\beta \end{bmatrix}e^{Dt}\begin{bmatrix} A\\ B \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0 & \alpha +\beta \end{bmatrix}\begin{bmatrix} 1 &0 \\ 0 & e^{(\alpha +\beta )t} \end{bmatrix}\begin{bmatrix} A\\ B \end{bmatrix}$

and $\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}=\begin{bmatrix} -A\frac{\beta }{\alpha } +Be^{(\alpha +\beta )t}\\ A+Be^{(\alpha +\beta )t} \end{bmatrix}$

Does any of this sound right? :/ Thank you for all the help!

 

[haruspex]

$\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}=\begin{bmatrix} -A\frac{\beta }{\alpha } +Be^{(\alpha +\beta )t}\\ A+Be^{(\alpha +\beta )t} \end{bmatrix}$

The system would have constant energy. You should have periodic trig functions in the answer, not exponentials. Probably a sign error.

 

[Kosomoko]

I think the problem is still with your expression for the force from the spring. The way you wrote it, being proportional to x₁+x₂, that would mean that if x₁ = x₂, the spring would be exerting a force. That's not right; if x₁ = x₂, then the spring is at its equilibrium length, so there should be no force from it.

p.s. I think your term for the force from gravity also has the wrong sign.

 

[skrat]

So for first body:

$-\ddot{\varphi _1}+\varphi _1(\frac{g}{l}+\frac{k}{m})+\varphi _2\frac{k}{m}=0$

and second:

$-\ddot{\varphi _2}+\varphi _2(\frac{g}{l}+\frac{k}{m})+\varphi _1\frac{k}{m}=0$

This gives me a system : $\begin{bmatrix} \ddot{\varphi _1}\\ \ddot{\varphi _2} \end{bmatrix}=\begin{bmatrix} \frac{g}{l}+\frac{k}{m} &\frac{k}{m}\\ \frac{g}{l}+\frac{k}{m} & \frac{k}{m} \end{bmatrix}\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}$

I made I mistake here... my system is $\begin{bmatrix} \ddot{\varphi _1}\\ \ddot{\varphi _2} \end{bmatrix}=\begin{bmatrix} \frac{g}{l}+\frac{k}{m} & \frac{k}{m}\\ \frac{k}{m}& \frac{g}{l}+\frac{k}{m} \end{bmatrix}\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}$

If $\alpha = \frac{g}{l}+\frac{k}{m}$ and $\beta = \frac{k}{m}$. We can see that the matrix is symmetrical $\begin{bmatrix} \alpha & \beta\\ \beta& \alpha \end{bmatrix}$

The eigen values are $(\alpha -\lambda) ^2-\beta ^2=\lambda ^2-2\alpha \lambda +(\alpha ^2-\beta ^2)$ which gives me $\lambda _{1,2}=\alpha \pm \beta$.

Accordingly, eigenvector for $\lambda _{1}=\alpha + \beta$ is $v_1=(1,1)$ and for $\lambda _{2}=\alpha - \beta$ a vector $v_2=(-1,1)$

This means, that matrix D from eigenvalues is $D=\begin{bmatrix} \alpha +\beta &0 \\ 0& \alpha -\beta \end{bmatrix}$ and matrix P (from eigenvectors) is $P=\begin{bmatrix} 1&-1 \\ 1& 1 \end{bmatrix}$.

The general solution should be like $\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}=\begin{bmatrix} 1&-1 \\ 1& 1 \end{bmatrix}\begin{bmatrix} e^{(\alpha +\beta )t} &0 \\ 0& e^{(\alpha -\beta )t} \end{bmatrix}\begin{bmatrix} A\\ B \end{bmatrix}$

so general solution is $\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}=\begin{bmatrix} Ae^{(\alpha +\beta )t}-Be^{(\alpha -\beta )t}\\ Ae^{(\alpha +\beta )t}+Be^{(\alpha -\beta )t} \end{bmatrix}$

Anything more I can do with that? That is, if it is not wrong again...

The system would have constant energy. You should have periodic trig functions in the answer, not exponentials. Probably a sign error.

Ok. How on can I check if the energy is constant? One more, since I have exponential functions again, my result is than again not ok?

I think the problem is still with your expression for the force from the spring. The way you wrote it, being proportional to x₁+x₂, that would mean that if x₁ = x₂, the spring would be exerting a force. That's not right; if x₁ = x₂, then the spring is at its equilibrium length, so there should be no force from it.

p.s. I think your term for the force from gravity also has the wrong sign.

Let this be the position at t=0. The total extensions of the spring at this point is $x_1+x_2$.
Unless I understood you wrong, if $x_1=x_2\neq 0$ the force of the spring will not be 0, which is ok.

Gravity has the same direction as the spring.

 

[Kosomoko]

Well, the usual way to set up the coordinates for this would be to have the positive x direction be the same for both masses. Meaning that in your diagram, mass 1 currently has a negative x position and mass 2 has a positive x position. What you are proposing involves instead defining a sort of opposite x-axis for the second mass; this may work, but it is certainly not the most simple thing to do.

Even accepting this, though, the signs of the terms in your equation are still not correct. Look, for example, at the term for gravity... you essentially have the force being proportional to phi. This will mean that if phi is, for example +1 degree, gravity would tend to increase it even more, resulting in an exponential increase, which is obviously wrong. It should instead be proportional to -phi, tending to pull the mass back to phi=0.

 

[haruspex]

$-\ddot{\varphi _1}+\varphi _1(\frac{g}{l}+\frac{k}{m})+\varphi _2\frac{k}{m}=0$

Shouldn't it be:
$\color{red}{+}\ddot{\varphi _1}+\varphi _1(\frac{g}{l}+\frac{k}{m\color{red}{l}})+\varphi _2\frac{k}{m\color{red}{l}}=0$?

 

[skrat]

First, to answer Kosomoko: I agree now. The total extension of the spring is $x_2-x_1$. Sorry for not believing you at first...

Shouldn't it be:
$\color{red}{+}\ddot{\varphi _1}+\varphi _1(\frac{g}{l}+\frac{k}{m\color{red}{l}})+\varphi _2\frac{k}{m\color{red}{l}}=0$?

Well... originally the conditions are $\sum M=J\ddot{\varphi }$

$J\ddot{\varphi _1}=F_glsin\varphi _1+F_slcos\varphi _1$

$\sum M=J\ddot{\varphi _1}=mglsin\varphi _1+k(x_2-x_1)lcos\varphi _1$

$J\ddot{\varphi _1}=mglsin\varphi _1+k(lsin\varphi _2-lsin\varphi _1)lcos\varphi _1$

For small $\varphi$ than $ml^2\ddot{\varphi }=mgl\varphi _1 +kl^2(\varphi _2-\varphi _1)$

O don't see how $\frac{k}{m\color{red}{l}}$ but bear in mind that looks like there are a lot of things i don't see..

 

[skrat]

$ml^2\ddot{\varphi }=mgl\varphi _1 +kl^2(\varphi _2-\varphi _1)$

Blah, I checked some old notes... And you guys are right...

it should be $-ml^2\ddot{\varphi }=mgl\varphi _1 +kl^2(\varphi _2-\varphi _1)$ allthough I don't really get it why but ok...

 

[haruspex]

O don't see how $\frac{k}{m\color{red}{l}}$

No, sorry - scratch that one. It is just k/m.
For the sign, you have the angles measured as positive when the masses are far apart. When > 0, both the gravitational force and the spring force act to reduce the angle.

 

[skrat]

$-\ddot{\varphi _1}=\frac{g}{l}\varphi _1 +\frac{k}{m}(\varphi _2-\varphi _1)$

and

$-\ddot{\varphi _2}=\frac{g}{l}\varphi _2 +\frac{k}{m}(\varphi _2-\varphi _1)$

give me

$\begin{bmatrix} \ddot{\varphi _1}\\ \ddot{\varphi _2} \end{bmatrix}=\begin{bmatrix} \frac{k}{m}-\frac{g}{l} & -\frac{k}{m}\\ \frac{k}{m}& -\frac{k}{m}-\frac{g}{l} \end{bmatrix}\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}$

where $\begin{bmatrix} \frac{k}{m}-\frac{g}{l} & -\frac{k}{m}\\ \frac{k}{m}& -\frac{k}{m}-\frac{g}{l} \end{bmatrix}$ has got eiganvalues $\lambda$ defined as $(\lambda +\frac{g}{l})^2=0$ What is interpretation of that?

 

[haruspex]

$-\ddot{\varphi _1}=\frac{g}{l}\varphi _1 +\frac{k}{m}(\varphi _2-\varphi _1)$

and

$-\ddot{\varphi _2}=\frac{g}{l}\varphi _2 +\frac{k}{m}(\varphi _2-\varphi _1)$

So you've switched to measuring the angles in the same direction now?
If so, you need to reverse the sign on one of the differences. E.g. if φ₁ is measured away from the common centre and φ₂ is measured towards the common centre then when φ₁ > φ₂ the spring will be under tension. That will act to reduce φ₁ but increase φ₂.
$-\ddot{\varphi _1}=\frac{g}{l}\varphi _1 -\frac{k}{m}(\varphi _2-\varphi _1)$
$-\ddot{\varphi _2}=\frac{g}{l}\varphi _2 +\frac{k}{m}(\varphi _2-\varphi _1)$

 

[skrat]

So you've switched to measuring the angles in the same direction now?
If so, you need to reverse the sign on one of the differences. E.g. if φ₁ is measured away from the common centre and φ₂ is measured towards the common centre then when φ₁ > φ₂ the spring will be under tension. That will act to reduce φ₁ but increase φ₂.
$-\ddot{\varphi _1}=\frac{g}{l}\varphi _1 -\frac{k}{m}(\varphi _2-\varphi _1)$
$-\ddot{\varphi _2}=\frac{g}{l}\varphi _2 +\frac{k}{m}(\varphi _2-\varphi _1)$

Yes, I've switched to positive angles.

$\begin{bmatrix} \ddot{\varphi _1}\\ \ddot{\varphi _2} \end{bmatrix}=\begin{bmatrix} -\frac{g}{l}-\frac{k}{m} & \frac{k}{m} \\ \frac{k}{m} & -\frac{g}{l}-\frac{k}{m} \end{bmatrix}$ for $\alpha =\frac{g}{l}$ and $\beta =\frac{k}{m}$ than $\begin{bmatrix} \ddot{\varphi _1}\\ \ddot{\varphi _2} \end{bmatrix}=\begin{bmatrix} -\alpha-\beta &\beta \\ \beta & -\alpha-\beta \end{bmatrix}$

Eigenvalues $\lambda ^2+\lambda (2\alpha+2\beta)+\alpha ^2+2\alpha \beta=0$ so $\lambda _{1,2}=\frac{-2(\alpha +\beta)\pm\sqrt{4\alpha ^2+8\alpha \beta +4\beta ^2-4\alpha ^2-8\alpha \beta }}{2}= -(\alpha + \beta) \pm \beta$

For $\lambda _1=-\alpha$ is eigenvector $v_1=(1,1)$ and for $\lambda _2=-\alpha-2\beta$ a vector $v_2=(-1,1)$

Alltogether this gives me
$\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}=\begin{bmatrix} Ae^{-(\alpha +2\beta)t}-Be^{-\alpha t}\\ Ae^{-(\alpha +2\beta)t}+Be^{-\alpha t} \end{bmatrix}$

I don't see any periodic functions here..:/

 

[Kosomoko]

α and β both have dimensions of s^-2. So the thing in the exponents of your answer has dimensions of s^-1, which is bad because it should be dimensionless.

You need to take the square root of the eigenvalues you found. To see why this is, it may be useful to express the vector (φ₁,φ₂) as a linear combination of the eigenvectors and rewrite the main differential equation(s) in terms of this. You will find that it decouples into two uncoupled differential equations whose solution is easy to see from looking at it.

 

[skrat]

Eigenvalues $\lambda ^2+\lambda (2\alpha+2\beta)+\alpha ^2+2\alpha \beta=0$ so $\lambda _{1,2}=\frac{-2(\alpha +\beta)\pm\sqrt{4\alpha ^2+8\alpha \beta +4\beta ^2-4\alpha ^2-8\alpha \beta }}{2}= -(\alpha + \beta) \pm \beta$

For $\lambda _1=-\alpha$ is eigenvector $v_1=(1,1)$ and for $\lambda _2=-\alpha-2\beta$ a vector $v_2=(-1,1)$

I can't see any square root of eigenvectors here if I honest with you. At least, that's how I was told to do it.

Don't get me wrong, it is obvious that the exponents have to be dimensionless - thanks for reminding me of that, but I can't and don't want to just write a square root there just because that's how it has to be.

Hmmm, I tried to follow your instructions and write $(\varphi _1, \varphi _2)$ using eigenvectors $v_1=(1,1)$ and $v_2=(-1,1)$... BUT... am... I don't quite understand you... Do you want me to rewrite these two equations:

$-\ddot{\varphi _1}=\frac{g}{l}\varphi _1 -\frac{k}{m}(\varphi _2-\varphi _1)$
$-\ddot{\varphi _2}=\frac{g}{l}\varphi _2 +\frac{k}{m}(\varphi _2-\varphi _1)$

or... ?

Btw: I imagined this would be an easy problem to solve... -.- Thanks to both for being patient!

 

[Kosomoko]

Yes, that is the equation I meant. For convenience, let...

\vec{\varphi} = \begin{bmatrix}\varphi_1 \\\varphi_2\end{bmatrix}

So right now, the equation you have is...

\ddot{\vec{\varphi}} = A\vec{\varphi}

Where A is that 2x2 matrix. Write φ as a sum of v₁ and v₂...

\vec{\varphi} = a\vec{v_1} + b\vec{v_2}

Insert that into the equation and you get...

\ddot{a}\vec{v_1} + \ddot{b}\vec{v_2} = A(a\vec{v_1} + b\vec{v_2}) = \lambda_1a\vec{v_1} + \lambda_2b\vec{v_2}

Since v₁ and v₂ are linearly independent, you can decouple that vector equation into two simple uncoupled differential equations for a and b...

\ddot{a} = \lambda_1a \\<br /> \ddot{b} = \lambda_2b

The solutions of those are pretty familiar. Now you have general solutions for a and b, you can put them back into this...

\vec{\varphi} = a\vec{v_1} + b\vec{v_2}

And now you have a general solution for φ.

 

[skrat]

Never in my life would I thought of that...

That gives me

$\begin{bmatrix} \varphi _1\\ \varphi _2 \end{bmatrix}=\begin{bmatrix} C_1e^{\sqrt{-\alpha }t}+C_2e^{-\sqrt{-\alpha }t}-C_3e^{\sqrt{-\alpha -2\beta }t}-C_4e^{-\sqrt{-\alpha -2\beta }t}\\ C_1e^{\sqrt{-\alpha }t}+C_2e^{-\sqrt{-\alpha }t}+C_3e^{\sqrt{-\alpha -2\beta }t}+C_4e^{-\sqrt{-\alpha -2\beta }t} \end{bmatrix}=\begin{bmatrix} C_1e^{i\sqrt{\alpha }t}+C_2e^{-i\sqrt{\alpha }t}-C_3e^{i\sqrt{\alpha +2\beta }t}-C_4e^{-i\sqrt{\alpha +2\beta }t}\\ C_1e^{i\sqrt{\alpha }t}+C_2e^{-i\sqrt{\alpha }t}+C_3e^{i\sqrt{\alpha +2\beta }t}+C_4e^{-i\sqrt{\alpha +2\beta }t} \end{bmatrix}$

 

[Kosomoko]

You can probably use some trigonometric identities to make it look nice and tidy. I think the eventual form should be something like sin(Ωt)sin(ωt) where Ω is a beat frequency, perhaps also with some phase shifts.

I would also recommend putting the solution back into the original equations to make sure it's right.