= Stationary middle mass in a three-mass spring chain
{title2=$(k_1+k_2)/m_1=k_3/m_3,\quad q_2=0,\quad q_3=-(k_2/k_3)q_1$}
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: $k_2q_1+k_3q_3=0$. The outer <equations of motion> then reduce to $m_1\ddot q_1=-(k_1+k_2)q_1$ and $m_3\ddot q_3=-k_3q_3$.
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.
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