How Do You Derive the Lagrangian for Three Coupled Pendulums?

[CNX]

Homework Statement

Three pendulums hand side-by-side and have there masses connected horizontally via springs. All lengths and masses are equal. Find the Lagrangian and put it in terms of "natural units".

The Attempt at a Solution

$$T = 1/2 m l^2 (\dot{\theta_1}^2 + \dot{\theta_2}^2 + \dot{\theta_3}^2)$$

$$V = 1/2 k (l \theta_1 - l \theta_2)^2 + 1/2 k (l \theta_2 - l \theta_3)^2$$

Using natural units $q_i = \sqrt{k} x_i$ and $q' = dq/d\tau$ where $\tau=\omega t$:

$$L = 1/2(q^{'2}_{1} + q^{'2}_{2} + q^{'2}_{3}) - 1/2(q^2_1 + q^2_2 + q^2_3) + (q_1 q_2 + q_2 q_3 - 1/2 q^2_2)$$

When I try to find the normal modes I only get 3, but since there are three degrees of freedom shouldn't there be three normal modes? I think my T or V expressions must be wrong somewhere. If they are not I can write out the matrices...

When there is no (1,3) or (3,1) entry in either the T or V matrix, i.e. this case (using natural units), will there ever be 3 modes when solving characteristic eq. for that matrix?

 

[queenofbabes]

When I try to find the normal modes I only get 3, but since there are three degrees of freedom shouldn't there be three normal modes?

Do you mean you only found one normal mode with $$\omega$$ = 3?

Your 3x3 matrix should have 3 eigenvalues and hence 3 normal modes, right?

 

[CNX]

No I mean I get $\omega^2_1 = 0,~\omega^2_2 = 3/m l^2, ~\omega^2_3 = 1/ m l^2$

From

$$-.375 m l^2 \omega+.500 m^2 l^4 \omega^2-.125 m^3 l^6 \omega^3=0$$

I don't really get the idea of the natural units. Is it just convenience, i.e. solve the eigenvalue problem and then convert back?

 

[queenofbabes]

To be honest I've already finished this topic but I've not heard of "natural units", but if you're solving for normal modes then what you've done looks correct...you have already found the 3 normal frequencies.

 

[CNX]

Thanks