Normal modes of two arc-spring-coupled beads on a hoop (${\omega }_0=0,{\omega }_1^2=2k(m_1+m_2)/(m_1{m}_2)$)

Two frictionless beads of positive masses $m_1,m_2$ on a circular hoop are joined by two identical springs of spring constant $k$, constrained to follow the two arcs. At their diametrically opposite equilibrium, each spring has natural length equal to half the circumference. For signed arc length displacements in the same rotational sense, the potential energy is $U=k(x_1-x_2{)}^2$. The normal mode of common rotation has mode shape $(1,1)$ and zero angular frequency, while the other normal mode has mode shape $(1,-m_1/m_2)$ and the positive angular frequency in the title.

Adding the equations of motion shows that $X=(m_1{x}_1+m_2{x}_2)/(m_1+m_2)$ satisfies $\ddot{X}=0$. Subtracting them gives $\ddot{\delta }+{\omega }_1^2\delta =0$ for $\delta =x_1-x_2$. Thus the complete zero mode is $X=C+Dt$, including uniform rotation, and the relative motion is a harmonic oscillator. In the limit $m_2/m_1\to \infty$ at fixed $m_1,k$, the oscillatory motion of the heavier bead vanishes for fixed lighter-bead amplitude, and ${\omega }_1\to \sqrt{2k/m_1}$. A common drift still survives if its initial velocity is held fixed.