# Mathemathical pendulum with springs

1. Apr 23, 2014

### skrat

1. The problem statement, all variables and given/known data
A body with mass $m$ is hanged on a line with length $l$ and attached to springs in point $p$. Point $p$ can move only horizontally. In equilibrium position, non of the springs is deformed. Now let's give that body just a little push out of equilibrium position. Calculate the frequency of oscillation.

2. Relevant equations

3. The attempt at a solution

Let's put the origin of our coordinate system in point $p$ with $\hat{i}$ axis to right along the right spring and $\hat{j}$ axis pointed up.

Now the coordinate is fixed, while point $p$ can move along axis $\hat{i}$. Let $\vec{r}$ be vector from the origin to mass $m$ and lets say that $\varphi$ is the angle between the line $l$ and vertical line.

Than $\vec{r}=(lsin\varphi + x,lcos\varphi )$ where $x$ is the expansion/shrinkage of springs or in other words: movement of point $p$.

Than $\dot{\vec{r}}=(\dot{\varphi }lcos\varphi+\dot{x},-l\dot{\varphi }sin\varphi)$ and $\dot{\vec{r}}^2=(\dot{\varphi }l)^2+2\dot{\varphi }l\dot{x}cos\varphi+\dot{x}^2$.

Now $L=T-V=\frac{1}{2}m\dot{\vec{r}}^2-(-mglcos\varphi +kx^2)=(\dot{\varphi }l)^2+2\dot{\varphi }l\dot{x}cos\varphi+\dot{x}^2+mglcos\varphi -kx^2$

If that is ok, than

$\frac{\partial L}{\partial \varphi }-\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{\varphi }}=\ddot{\varphi }+\frac{g}{l}sin\varphi +\frac{\ddot{x}}{l}cos\varphi =0$ and

$\frac{\partial L}{\partial x}-\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{x }}=\ddot{x}+\frac{2k}{m}x+\ddot{\varphi }lcos\varphi -\dot{\varphi }^2lsin\varphi =0$

Hopefully so far everything is ok. Now I am lost here. I am somehow supposed to solve this system of differential equations yet I have no idea how. I tried applying Taylor expansion for small $\varphi$ but that didn't really help much...

2. Apr 24, 2014

### vanhees71

Taylor expanding for small $\varphi$ and $\dot{\varphi}$ helps enormously. Take the expansion up to linear order and you get a set of two coupled linear equations of motion which you can solve with standard methods, finding the eigensystem of solutions.

3. Apr 24, 2014

### skrat

Yup, I figured it out yesterday late into the night...

For small angles:

$\ddot{\varphi }+\frac{g}{l}\varphi +\frac{1}{l}\ddot{x}=0$ and

$\ddot{x}+\frac{2k}{m}x +l\ddot{\varphi }=0$

Now lets say that $x=x_0e^{i\omega t}$ and $\varphi =\varphi _0e^{i\omega t}$. This gives me

$\begin{bmatrix} -\omega ^2+\frac{g}{l} & -\omega ^2\\ -\omega ^2& -\omega^2 +\frac{2k}{m} \end{bmatrix}\begin{bmatrix} \varphi _0\\ x_0 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

And finally $\omega ^2=\frac{g/l}{1+\frac{mg}{2kl}}$

THANKS!

4. Apr 24, 2014

### vanhees71

Hm, aren't there two resonance frequencies? You must have a solution with four free parameters, determined by the initial conditions $x(0)=x_0$, $\dot{x}_0=v_0$, $\varphi(0)=\varphi_0$, and $\dot{\varphi}(0)=\Omega_0$.

5. Apr 24, 2014

### skrat

I don't know. I am not smart enough, that's why I am here learning. :D You tell me?

I do have some unofficial solutions for this problem and my solution is the same. That's all I know.

Maybe I wasn't clear enough when writing the problem: It says here that we grab the mass and pull it away from equilibrium position for small $\varphi$ and than release it. Does this change anything?