For an impulse vector delivered at an unstrained equilibrium, project the velocity jump onto mass-orthogonal normal modes. Multiplying by gives modal velocity coefficient . Its kinetic energy is the displayed expression, including a translational zero mode if one is present. A single endpoint kick in an equal-mass free three-mass spring chain puts fractions into translation and the two internal modes.
A normal mode is a pattern in which all coordinates share one frequency and fixed relative amplitudes. For a conservative linear system, it solves a generalized eigenvalue equation with stiffness matrix and mass matrix . For this free three-mass spring chain,
Solving , for , gives
Their overall normalization is arbitrary; distinct modes are orthogonal in the mass inner product. The zero mode is rigid translation because neither end is fixed. In the antisymmetric mode the middle mass stays still, so changing its mass cannot alter the frequency. The symmetric internal mode keeps the centre of mass fixed through its middle displacement . Its frequency rises without bound as and approaches as , becoming degenerate with the antisymmetric frequency in that limit.
For equal masses put and . Positions are equilibrium positions plus displacements
The translational mode needs the linear-in-time term; a cosine at zero frequency alone would miss arbitrary centre-of-mass velocity.
Taking the springs initially at equilibrium, the impulse gives . Their squared Euclidean norms are , so their initial kinetic energies are . Relative to total ,
Subsequent motion conserves each mode's total energy, exchanging its kinetic and spring contributions. The zero-mode share remains centre-of-mass kinetic energy.
Figure 1.
Original displacement-vector sketches and mass-ratio dependence of the free three-mass chain modes
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