Taking downward distance as positive, the Lagrangian is

L = \frac{1}{2} (3 m) \overset{z}{˙}_{1}^{2} + \frac{1}{2} (2 m) \overset{z}{˙}_{2}^{2} - \frac{k}{2} (z_{1} - l)^{2} - \frac{k}{2} (z_{2} - z_{1} - l)^{2} + 3 m g z_{1} + 2 m g z_{2} .

At equilibrium the lower spring supports only the lower mass, while the upper spring supports both masses:

k (z_{2} - z_{1} - l) = 2 m g, k (z_{1} - l) = 5 m g .

Hence

z_{1}^{(0)} = l + \frac{5 m g}{k}, z_{2}^{(0)} = 2 l + \frac{7 m g}{k} .

Let

q_{i} = z_{i} - z_{i}^{(0)}

. The linear terms cancel by the equilibrium equations, and the quadratic Lagrangian is

L = \frac{1}{2} \dot{q}^{T} T \dot{q} - \frac{1}{2} q^{T} V q + constant,

T = m (3002), V = k (2 - 1 - 1 1) .

(V - ω^{2} T) a = 0.

Its characteristic equation is

det (V - ω^{2} T) = k^{2} - 7 km ω^{2} + 6 m^{2} ω^{4} = 0,

ω_{1} = \frac{k}{6 m}, ω_{2} = \frac{k}{m} .

a_{1} = (2, 3)^{T}, a_{2} = (1, - 1)^{T} .

They are not orthogonal in the ordinary Euclidean inner product. They are orthogonal in the kinetic-energy, or mass-matrix, inner product:

a_{1}^{T} T a_{2} = (2, 3) m (3002) (- 1 1) = 0 .

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 Incoming links: Mass-matrix orthogonality of normal modes