Stationary middle mass in a three-mass spring chain ($(k_1+k_2)/m_1=k_3/m_3,q_2=0,q_3=-(k_2/k_3)q_1$)

Consider a straight chain with a fixed wall, spring constants $k_1,k_2,k_3>0$, and consecutive point masses $m_1,m_2,m_3>0$, with the third mass freely ending the chain. Displacements are measured from equilibrium. If the middle mass stays at equilibrium, the two adjacent spring forces must cancel: $k_2{q}_1+k_3{q}_3=0$. The outer equations of motion then reduce to $m_1{\ddot{q}}_1=-(k_1+k_2)q_1$ and $m_3{\ddot{q}}_3=-k_3{q}_3$.

A nonzero oscillatory normal mode with $q_2=0$ therefore exists exactly when the two outer squared angular frequencies agree, as displayed. Its mode shape is proportional to $(1,0,-k_2/k_3)$ and its frequency does not depend on $m_2$. Appropriate initial displacements and velocities confined to this shape keep the middle mass still. For $k_1=2k$, $k_2=k_3=k$, $m_1=3m$, $m_3=m$, the common frequency is $\sqrt{k/m}$ and the outer displacements are equal and opposite.