# Small oscillations+normal modes of a system

1. Feb 20, 2012

### fluidistic

1. The problem statement, all variables and given/known data
Two identical pendulums of length $l$ hang from a ceiling. Their vertical axis is separated by a distance $l_0$. They are made by 2 masses m. Between these 2 masses we put a spring of constant k and natural length $l_0$.
Gravity acts verticaly downward.
1)Calculate the proper frequencies of oscillations in the approximation of small oscillations.
2)Determine the normal coordinates of oscillations.
3)Suppose that at $t=0$ both pendulums are at rest for $\theta _1=0$ and $\theta _2=\theta _0$; find and solve the equations of motion of the system.
Note that $\theta _1$ and $\theta _2$ are the angles that each mass makes with the vertical.
2. Relevant equations
$L=T-V$.

3. The attempt at a solution
I tried to get the Lagrangian of the system which consists of the sum of 3 Lagrangians, namely one for each mass and one for the spring.
I'm having trouble with the potential energy stored in the spring.
Let $L_1$ and $L_2$ be the Lagrangians for each mass. I found out that $T_1 =\frac{m}{2}l^2 \dot \theta _1 ^2$ while $T_2 =\frac{m}{2}l^2 \dot \theta _2 ^2$. Potential energy is simply $V_1 = mgl [\cos (\theta _1 ) -1]$ and $V_2 = mgl [\cos (\theta _2 ) -1]$. So I can just make up the 2 Lagrangians for these masses.
Now the hard part: for the spring... $L_3=V_3$.
$V_3=\frac{k}{2}(\Delta x ) ^2$ where delta x denotes the elongation/compression of the spring with respect to $l_0$. So if $\vec r_1$ and $\vec r_2$ denotes the positions of mass 1 and mass 2 respectively, I have that $\Delta x = ||\vec r_1 - \vec r_2|-l _0|$. Since I will consider $(\Delta x ) ^2$ I can safely ignore the external "modulus" ||.
Anyway here is my work: I take my reference system origin at the position of the mass 1 when nothing moves (stable equilibrium position), I have that $\vec r_1 = l\sin \theta _1 \hat i+l (1-\cos \theta _1) \hat j$ and $\vec r _2 =(l_0+l \sin \theta _2)\hat i + (l-l \cos \theta _2 ) \hat j$.
Thus $\vec r_2 -\vec r_1 =[l _0-l (\sin \theta _2 + \sin \theta _1 )]\hat i +l (\cos \theta _1 - \cos \theta _2 )\hat j$.
Now I used some trig identities at http://www.sosmath.com/trig/Trig5/trig5/trig5.html to get that $\vec r_2 -\vec r_1 =\{ l_0-l \left [ 2 \sin \left ( \frac{\theta _1+ \theta _2 }{2} \right ) \cos \left ( \frac{\theta _1- \theta _2 }{2} \right ) \right ] \} \hat i+l \left [ -2 \sin \left ( \frac{\theta _1+ \theta _2 }{2} \right ) \sin \left ( \frac{\theta _1- \theta _2 }{2} \right ) \right ] \hat j$.
I thus calculated $|\vec r_2 - \vec r_1 |=\sqrt {\left [ l_0-2l \sin \left ( \frac{\theta _1 +\theta _2 }{2} \right ) \cos \left ( \frac{\theta _1 -\theta _2 }{2} \right ) \right ]^2+\left [ 2l \sin \left ( \frac{\theta _1 +\theta _2 }{2} \right ) \sin \left ( \frac{\theta _1 -\theta _2 }{2} \right ) \right ]^2}$. I simplified this to $\sqrt {l_0 ^2 -4ll_0 \sin \left (\frac{\theta _1 + \theta _2 }{2} \right )\cos \left (\frac{\theta _1 - \theta _2 }{2} \right ) +4 l^2\sin ^2 \left (\frac{\theta _1 + \theta _2 }{2} \right ) }$.
Despite having checked out 3 times the algebra I cannot find any mistake, yet the result makes no sense. Indeed, for when $\theta _1 = \theta _2$, I should get $|\vec r_2 - \vec r_1 |=l_0$ but I get a condition for this to be true instead of it being true no matter what. I get the condition $l_0 \sin \theta =l$ for all angles theta. Since this isn't necessarily true, something is wrong... but I don't see what.
I'd appreciate any help and would be glad to know where my mistake lies. Great thanks in advance.

2. Feb 20, 2012

### fluidistic

I just spotted an algebra mistake for $\vec r_2 -\vec r_1$. Working on that!

3. Feb 27, 2012

### fluidistic

I get $L=\frac{m}{2}l^2(\dot \theta _1 ^2+\dot \theta _2^2)+mgl(2-\cos \theta _1 - \cos \theta _2 )-\frac{k}{2} \{ l_0^2+2ll_0 (\sin \theta _2 - \sin \theta _1)+2l^2[1-\cos (\theta _1-\theta _2 )] \}$.
So that $V(\theta _1 , \theta _2 )=mgl[\cos \theta _1 + \cos ( \theta _2 ) -2]+\frac{k}{2} \{ l_0^2+2ll_0 (\sin \theta _2 - \sin \theta _1)+2l^2[1-\cos (\theta _1-\theta _2 )] \}$.
Now I am unsure if I must use the small angles approximation already, or first approximate this potential by a function of the form $\frac{k}{2}x^2$ in which case I do not know how to do. Any help is appreciated.
Edit: Just spotted another algebra mistake, the potential energy should be even more complicated...

Edit 2:Considering from start that $\sin \theta \approx \theta$ and $1-\cos \theta \approx \frac{\theta ^2}{2}$ I reach $V\approx \frac{mgl}{2} (\theta _1 ^2 + \theta _2 ^2 )+k[l_0 ^2+ll_0(\theta _2 - \theta _1 )+\frac{l^2(\theta _2-\theta _1 )^2}{2}-l_0 \sqrt {l_0 ^2+2ll_0 (\theta _2-\theta _1 )+l^2(\theta_2 - \theta _1)^2}]$. Could someone confirm this?

Last edited: Feb 27, 2012
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