Solution

For equal masses and springs,

L_{1} = ℓ + \frac{2 m g}{k}, L_{2} = ℓ + \frac{m g}{k} .

A small horizontal displacement

x_{1}

changes each spring direction to first order. The equilibrium tensions

2 m g

and

m g

give transverse restoring stiffness

K_{x} = \frac{2 m g}{L _{1}} + \frac{m g}{L _{2}} .

Therefore

ω_{1}^{2} = \frac{K _{x}}{m} = \frac{k}{m} (2 - \frac{1}{1 + 2 m g / ( k ℓ )} - \frac{1}{1 + m g / ( k ℓ )}),

as required.

The other motions are vertical. If

η_{1}, η_{2}

are downward displacements from equilibrium, their linearized equations are

m (\overset{η}{¨}_{1} \overset{η}{¨}_{2}) = - k (2 - 1 - 1 1) (η_{1} η_{2}) .

ω_{\pm}^{2} = \frac{k}{2 m} (3 \pm 5), (η_{1}, η_{2}) \propto (1, \frac{1 \mp 5}{2}) .

Neither frequency contains

g

: gravity only shifted the equilibrium lengths.

Solved by gpt-5.6-sol high.

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