Discussion
In the eigenvalue problem \( (K - \omega^2 M)\phi = 0 \), the non-trivial solutions require that
Answer: B
Correct answer: the determinant of \( K - \omega^2 M \) vanish
The equation is homogeneous, so a non-zero \( \phi \) exists only if the matrix is singular:
\[ \det(K - \omega^2 M) = 0 \]
Expanding the determinant gives the characteristic polynomial in \( \omega^2 \), whose \( n \) roots are the natural frequencies. Substituting each root back gives the corresponding mode shape, determined only up to a scale factor - which is why mode shapes are normalised rather than absolute.
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