# Small Osc. Pendulum+Springs

1. Jul 16, 2009

### CNX

1. The problem statement, all variables and given/known data

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".

3. 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?

Last edited: Jul 16, 2009
2. Jul 16, 2009

### queenofbabes

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?

3. Jul 16, 2009

### 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?

Last edited: Jul 16, 2009
4. Jul 16, 2009

### 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.

5. Jul 17, 2009

Thanks