 Octave tuning in a two-mass complete spring graph

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.