Solution

Subtracting the two linearized equations eliminates their acceleration terms:

ω_{0}^{2} θ = ω_{1}^{2} η .

For

g, k > 0

, the assumed nontrivial simple harmonic motion therefore has

η (t) = \frac{ω _{0}^{2}}{ω _{1}^{2}} θ_{0} sin (ω t) = \frac{m g}{k ℓ} θ_{0} sin (ω t) .

The support displacement and pendulum angle oscillate in phase. Inserting this relation into either Euler-Lagrange equation yields

ω^{2} (1 + \frac{ω _{0}^{2}}{ω _{1}^{2}}) = ω_{0}^{2},

so the one positive finite angular frequency is

ω = \frac{ω _{0} ω _{1}}{ω _{0}^{2} + ω _{1}^{2}} = \frac{g k}{m g + k ℓ}, \frac{1}{ω ^{2}} = \frac{ℓ}{g} + \frac{m}{k} .

The negative frequency describes the same real oscillation after a phase change. For a general phase, replace the sine in both coordinates by

A cos (ω t) + B sin (ω t)

There is no missing second finite branch. The quadratic Lagrangian is

L_{2} = \frac{1}{2} (\overset{η}{˙} + \dot{θ})^{2} - \frac{1}{2} ω_{1}^{2} η^{2} - \frac{1}{2} ω_{0}^{2} θ^{2},

whose kinetic matrix is

(1111)

and has rank one. Its characteristic determinant is

ω_{0}^{2} ω_{1}^{2} - (ω_{0}^{2} + ω_{1}^{2}) ω^{2}

ω^{2}

. This singular kinetic matrix in small oscillations results from taking the support mass to be exactly zero. Adding a small support mass restores a second mode that runs to infinite frequency as that mass vanishes.

For a stiff spring,

k \to \infty : η \to 0, ω \to g / ℓ .

k

, horizontal support motion makes the frequency smaller.

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