Finding Normal Modes in Coupled Oscillations

AI Thread Summary
The discussion focuses on deriving the normal modes of oscillation for two equal masses connected by springs. The frequencies of the normal modes are expressed as ω² = (3 ± √5)(s/2m), with specific amplitude ratios for the upper and lower masses in each mode. The user is struggling with the equations of motion and the subsequent calculations to isolate ω². Guidance is provided to expand the determinant equation into a quadratic form, allowing for the application of the quadratic formula to solve for ω². The conversation emphasizes the importance of correctly setting up and manipulating the equations to find the desired results.
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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
 
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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.
 
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
 
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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.
 
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