Two pendula connected by a spring - normal modes

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SUMMARY

The discussion centers on the equations of motion for two pendula connected by a spring, represented as m(d²x₁/dt²) = -(mg/l)x₁ + k(x₂-x₁) and m(d²x₂/dt²) = -(mg/l)x₂ + k(x₁-x₂). The stiffness matrix K is identified as K = [[mg/l + k, -k], [-k, mg/l + k]]. The substitution of x₁ = A₁e^(iωt) and x₂ = A₂e^(iωt) leads to the angular frequency ω = √(mg/l). The user seeks clarification on determining the eigenvalues and eigenvectors of the matrix K.

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Homework Statement
Two identical pendula each of length l and with bobs of mass m are free to oscillate in the same plane. The bobs are joined by a massless spring with a small spring constant k, such that the tension in the spring is k times its extension. Find the equations governing the two bobs, and write them in the form ##m\frac{d^2x}{dt^2} = -Kx## and write down the K matrix. Subsitute a normal mode solution ##x=af((t)# and show that this satisifes the equation of motion if a is an eigenvector. Solve the corresponding equation for f(t). How many eigenvectors does K have? Write them and find a general solution. At t=0, both pendula are at rest with x1=x2=a. Describe the subsequent motion of the two pendula.
Relevant Equations
##Ae^iwt##
##ma=m\frac{d^2x}{dt^2}##
I found the equations of motion as

##m\frac{\mathrm{d}^2x_1 }{\mathrm{d} t^2} = -\frac{mg}{l}x_1 + k(x_2-x_1)##

and
##m\frac{\mathrm{d}^2x_2 }{\mathrm{d} t^2} = -\frac{mg}{l}x_2 + k(x_1-x_2)##

I think the k matrix might be
##\begin{bmatrix}
mg/l + k & -k \\
-k & mg/l + k
\end{bmatrix}##
Then when I substitute
##x_1 = A_1e^(i\omega t)##
and
##x_2 = A_2e^(i\omega t)##
and found ##\omega = sqrt{\frac{mg}{l}}##
But how do I do the rest?
 
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Your result for omega has the wrong units.

What are the eigenvalues and eigenvectors of K?
 

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