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For two masses, connect each to a fixed anchor and to the other mass with equal stiffness $k$. Set $m_1=1,k=1$ to choose units. The squared angular frequencies have characteristic equation $m_2{z}^2-2(m_2+1)z+3=0$. Requiring a frequency ratio of two means that these roots are $z$ and $4z$. Their sum and product give $16m_2^2-43m_2+16=0$, yielding the two reciprocal ratios in the header. Restoring units multiplies all squared frequencies by $k/m_1$, leaving the permitted mass ratios unchanged. The term octave refers here only to this two-to-one ratio.

Release from rest gives $B=D=0$. At $t=0$, $A+2C=d$ and $A-C=0$, so $A=C=d/3$. ConsequentlyThe frequency ratio is exactly two, so both motions repeat after $2\pi /{\omega }_{-}$. These are commensurate normal modes, rather than a pair of nearly equal frequencies producing slow beats.

Let $z={\omega }_1^2$, so ${\omega }_2^2=4z$. The sum and product of roots of the characteristic polynomial in (i) giveEliminating $z$ yieldsTherefore the octave tuning in a two-mass complete spring graph has two positive solutions:The two values are reciprocals. Both meet the requested frequency ratio; substituting either into the expressions in (i) gives ${\omega }_2/{\omega }_1=2$.