Finding Normal Modes in Coupled Oscillations

[blackbody]

Hi guys, I'm stuck on a problem that states:

Two equal masses oscillate in the vertical direction. Show that the frequences of the normal modes of oscillation are given by:

$$\omega^2 = (3 +- \sqrt{5})\frac{s}{2m}$$

and that in the slower mode the ratio of the amplitude of the upper mass to that of the lower mass is $$\frac{1}{2}(\sqrt{5}-1)$$ whilst in the faster mode this ratio is $$\frac{-1}{2}(\sqrt{5}+1)$$. The figure in the problem is basically:

______
s
s
s
M
s
s
s
M

Where s is the spring and M are the masses (both springs have equal stiffness s).Basically, I'm not sure if I have the equations of motion down correctly. So far, I have:

1) $$\frac{md^2y_{1}}{dt^2} = -sy_{1} + s(y_{2} - y_{1}) => y_{1}'' = -\omega_{0}^2y_{1} + \omega_{0}^2(y_{2} - y_{1})$$

2) $$\frac{md^2y_{2}}{dt^2} = -s(y_{2} - y_{1}) => y_{2}'' = -\omega_{0}^2(y_{2} - y_{1})$$

Where $$y_{1}$$ and $$y_{2}$$ are the displacements of the first and second mass, respectively.
However when I use the solutions $$y_{1} = A_{1}\cos{wt}$$ and $$y_{2} = A_{2}\cos{wt}$$, find the derivatives, plug back in, etc, I cannot cleanly solve for the normal modes in terms of $$\omega$$. I'm suspecting my equations of motion are incorrect, help?

Thanks

 

[jamesrc]

Can you show some of your work so that we can see where you might be getting hung up? Your equations of motion appear fine.

 

[blackbody]

1st equation: $$-A_{1}w^2\cos{wt} = -w_{0}^2A_{1}\cos{wt} + w_{0}^2(A_{2}\cos{wt} - A_{1}\cos{wt})$$

cosines factor out, collect like terms, etc...then I get:

$$A_{1}(2w_{0}^2 - w^2) + A_{2}(-w_{0} = 0$$

2nd equation: $$-A_{2}w^2\cos{wt} = -w_{0}^2(A_{2}\cos{wt} - A_{1}\cos{wt})$$

cosines factor out, collect like terms, etc...then I get:

$$A_{1}(-w_{0}^2) + A_{2}(w_{0}^2 - w^2) = 0$$

So finally I end up with a system of equations:

$$A_{1}(2w_{0}^2 - w^2) + A_{2}(-w_{0}) = 0$$

$$A_{1}(-w_{0}^2) + A_{2}(w_{0}^2 - w^2) = 0$$

To tackle this, I set the determinant of the matrix equal to zero:

$$(2w_{0}^2 - w^2)(w_{0}^2 - w^2) - (w_{0}^2)^2 = 0$$

From this, I can't isolate $$w^2$$ to get the answer.

Thanks

 

[jamesrc]

Expand your last equation and collect like terms. The equation will then be in the form: $$a\omega^4+b\omega^2+c=0$$ where a, b, and c are constants. You can solve for $$\omega^2$$ using the quadratic equation.