# A stick with two springs attached between two vertical walls

1. Jun 25, 2014

### skrat

1. The problem statement, all variables and given/known data
A stick with mass $m=0.1kg$ and length $a=0.5m$ has on each end welded a spring with constant $k=2000N/m$ and length $l=0.2m$. We stretch both of the springs and mount them on the sides of vertical walls which are $d=1m$ apart. Find both translational and rotational eigen frequency for small oscillations in horizontal plane.

Deflection of the stick due to it's mass is negligible.

2. Relevant equations

3. The attempt at a solution

And here we go again. Another problem I can't solve by myself.

I started writing dynamic equations for each direction separately. Since everything happens in the $xy$ plane I started with $x$ direction, where the origin is of course in the center of mass of the rod when in equilibrium position.

So $m\ddot x=-F_1cos\vartheta -F_2cos\varphi$

$m\ddot x= -ks_1cos\vartheta -ks_2cos\varphi$

Where $s=s_0+\delta s$ is the stretch and $s_0=d/2-a/2-l$ initial deformation.

$s_1=s_0+(\sqrt{(l+x)^2+y^2}-s_0)=\sqrt{(l+x)^2+y^2}$

$s_2=s_0+(s_0-\sqrt{(l-x)^2+y^2})=2s_0-\sqrt{(l-x)^2+y^2}$

If I insert that into dynamic equation and also write cos in terms of x :

$cos\vartheta =\frac{d/2-a/2+x}{\sqrt{(l+x)^2+y^2}}$ and $cos\varphi =\frac{d/2-a/2-x}{\sqrt{(l-x)^2+y^2}}$

Now If I am not completely mistaken, this brings my dynamic equation to:

$m\ddot x= -2kx-\frac{ks_0}{l}(d/2-a/2)-\frac{ks_0x}{l}(d/2-a/2)$

The same idea goes for $y$ direction where I get

$m\ddot y =-\frac{2ks_0}{l}y$

All this must be terribly wrong. :D

2. Jun 25, 2014

### BvU

As you say: Here we go again! By now you mus be a real expert on springs!
The one translational freq is close to trivial, the other has links with a lengthy earlier post. So does the rotational one. I do miss the spring mass here, even though 0.1 kg for the stick isn't all that much. Shall we assume ideal springs ? Do we want to make life easy and assume stick and springs make up a straight line ? Or do we allow some sagging, thereby effectively excludng rotations in the xy plane ? I vote for the easy way...

Now you have done a lot already and you're definitely not dumb and no beginner. So why this "terribly wrong" fun spoiler?

01:50 here. Will check in tomorrow and I really look forward diving into this one !

3. Jun 25, 2014

### Staff: Mentor

This problem statement is a little confusing, but I'm going to assume that all the motion is taking place in the horizontal plane. The key feature of this problem is that there are three displacement degrees of freedom, not just one. Two of these degrees of freedom are the x and y displacements of the center of mass of the rod relative to the equilibrium position, and the other degree of freedom is the angle that the rod makes with the x axis. Once these three parameters are specified, the locations of all points along the rod are determined, including the ends of the rod. Once we know the locations of the ends of the rod, we then know the forces that the springs exert on each of the two ends (including their directions).

Chet

4. Jun 26, 2014

### skrat

Yes, springs are ideal. And yes, life is complicated enough, so let's just put everything up a straight line. :D

Well I expected $x$ and $y$ direction to be somehow coupled and not completely independent.

I think your assumption is ok since the original problem says: Find both translational and rotational eigen frequency for small oscillations in horizontal plane.

Ok, I originally meant to do that too - to write my dynamic equations with three degrees of freedom. However, the sin and cos projections can be directly expressed in terms of x and y, so I thought to myself.... Why would I have three degrees of freedom than?

5. Jun 26, 2014

### skrat

Ok, forget about my first attempt to solve the problem. It is probably fine but I ignored the fact that the stick can also rotate like chet warned me.

So I tried once again and I got an awesome expression for the stretch of the spring:

$s=s_0+\delta s$ where $s_0=d/2-a/2-l$, here $a$ is the length of the stick and $l$ the length of undeformed spring while $d$ is the distance between the vertical walls.

Now $\delta s= \sqrt{(\frac a 2 sin(\varphi + \vartheta)+y)^2+(\frac d 2 -\frac a 2-l+x +\frac a 2 cos(\varphi + \vartheta))^2}$

:D

$\varphi$ is the angle on the left side between the spring and x axis while $\vartheta$ is on the right side.

6. Jun 26, 2014

### Staff: Mentor

I came up with something different. If the center of mass of the spring is initially at the origin (0,0), then the springs are attached at (-d/2,0) and (+d/2,0), and the ends of the rod are at (-a/2,0) and (+a/2,0). If the center of mass is displaced Δx, Δy, and the spring rotates by an angle θ relative to the x axis, then the new coordinates of the ends of the rod are going to be at $(Δx-\frac{a}{2}cosθ,Δy-\frac{a}{2}sinθ)$ and $(Δx+\frac{a}{2}cosθ,Δy+\frac{a}{2}sinθ)$. The length of the left spring will now be:
$$L_l^2=(Δx-\frac{a}{2}cosθ+\frac{d}{2})^2+(Δy-\frac{a}{2}sinθ)^2$$
The length of the right spring will now be:
$$L_r^2=(-Δx-\frac{a}{2}cosθ+\frac{d}{2})^2+(Δy+\frac{a}{2}sinθ)^2$$

Chet

7. Jun 26, 2014

### BvU

Skrat, does your (currently empty ) set of relevant equations also include the Euler-Lagrange equations? Or do you have to make do with the Newton eqns and F= -kΔx and $\ddot x= -\omega^2x\,$, so to say the more introductory physics stuff ?

8. Jun 26, 2014

### Staff: Mentor

Hi Scrat,

This refers to my post #6. To linear terms in Δx, Δy, and θ, what are the lengths of the left- and right springs Ll and Lr, respectively? If Tr represents the tension in the right spring, to linear terms in Δx, Δy, and θ, what are the x- and y components of Tr? Same for the tension in the left spring.

Chet

9. Jun 27, 2014

### skrat

In general it includes everything, also Euler-Lagrange equations. However the idea is to solve this problem using Newton dynamic equations.

Chet, I will post an answer to your hints shortly. I tried to do some calculations already but everything looks horrible already for one direction only. I will do that, but right now I need to take some minutes to relax my brain.

10. Jun 27, 2014

### Staff: Mentor

Hi Skrat. When you linearize the force and moment balances with respect to the three degrees of freedom Δx, Δy, and θ, you will find that the equations become are much, much, much, much simpler. You will end up with 3 linear ODEs, and the equations will be uncoupled.

Chet