A positive-eigenvalue normal mode satisfies . Solving the constant-coefficient equation gives the sine, cosine, hyperbolic sine and hyperbolic cosine family. To avoid confusing modal coefficients with the filament bending modulus, write those coefficients as .
The free conditions at zero require and . With , the remaining conditions are
Its determinant is . Therefore the free-free bending spectrum is
A convenient corresponding eigenfunction is
where gives unit norm. The stable characteristic equation for free-free bending modes is . Intersections or numerical bracketing give
For the entire positive sequence, let . Expanding near its zero and using gives
There are also two zero-energy filament modes, not captured by substituting into the finite-coefficient trigonometric expression. Solving with free endpoints gives the affine kernel. An orthonormal basis of that kernel is
The positive normal modes are orthogonal to both. Including these two rigid-motion modes is essential for completeness; the regular self-adjoint operator on a finite interval has a complete discrete eigenbasis.
Figure 1.
Characteristic roots and bending shapes of a filament with two free ends
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