Normal modes, 2 masses, 2 springs

[Hume Howe]

Homework Statement

hey, this is my first time using this forum and was wondering if i could have some help with this normal modes question.

suspended from a ceiling is in order: a spring of spring constant (k_{2}), a mass (m), a spring with spring constant (k_{1}), and another mass (m)

calulate the normal mode frequencies \alpha_{1}, \alpha_{1} sorry can't find the omega symbol.

attempted solution:
let the displacement of the higher mass be x and the displacement of the lower mass be y.

m\ddot{x}=-k_{2}x + k_{1}(y-x)
m\ddot{y}=-k_{1}(y-x)

\left| \alpha^{2} - (\frac{k_{2}}{m} + \frac{k_{1}}{m}) -----\frac{k_{1}}{m} \right|
\left| \frac{k_{1}}{m} -------- \alpha^{2} - \frac{k_{1}}{m} \right| = 0

determinant = 0

so \alpha^{2} = 1/2[ (\frac{k_{2}}{m}+ \frac{2k_{1}}{m}) +- sqrt[(\frac{k_{2}}{m}+ \frac{2k_{1}}{m})^2 – 4(\frac{k_{2}*k_{1}}{m^2})]]

after simplifying it doesn't work when I use to try and solve for the amplitude ratios.
please advise, thanks.
p.s. is there a simpler equation writer i could download and use here or should i just persevere and try to use this embedded one?

 

[member 553083]

Homework Equations m*\ddot{x} = -k_{2}*x + k_{1}(y-x)m*\ddot{y} = -k_{1}(y-x)The Attempt at a Solution let the displacement of the higher mass be x and the displacement of the lower mass be y.m\ddot{x}=-k_{2}x + k_{1}(y-x)m\ddot{y}=-k_{1}(y-x)\left| \alpha^{2} - (\frac{k_{2}}{m} + \frac{k_{1}}{m}) -----\frac{k_{1}}{m} \right|\left| \frac{k_{1}}{m} -------- \alpha^{2} - \frac{k_{1}}{m} \right| = 0determinant = 0so \alpha^{2} = 1/2[ (\frac{k_{2}}{m}+ \frac{2k_{1}}{m}) +- sqrt[(\frac{k_{2}}{m}+ \frac{2k_{1}}{m})^2 – 4(\frac{k_{2}*k_{1}}{m^2})]]\alpha_{1,2} = 1/2[ (\frac{k_{2}}{m}+ \frac{2k_{1}}{m}) +- sqrt[(\frac{k_{2}}{m}+ \frac{2k_{1}}{m})^2 – 4(\frac{k_{2}*k_{1}}{m^2})]]^{\frac{1}{2}}