 Solution

For the stated geometry and numerical masses, the inertia and stiffness matrices areA normal mode $\mathbf{\theta }=\mathbf{v}{e}^{i\omega t}$ requires $(K-{\omega }^{2}M)\mathbf{v}=0$. Writing $z={\omega }^2$, its characteristic equation isAt the claimed $z=6$, this determinant is $-27/50\ne 0$. Hence the printed claim that $\sqrt{6}$ is a normal frequency is false for the stated data, and there is no corresponding nonzero normal mode.

The complete geometry-based answer is obtained from the three positive roots $z_j$:An exact closed form for the squared angular frequencies isThey are approximately $19.555611657$, $6.209018836$ and $1.235369507$, respectively. For any of these roots, a corresponding eigenvector can be chosen asThe last two rows of $(K-zM)\mathbf{v}=0$ give this expression directly; the first is the characteristic equation. These three modes span all motion, with arbitrary sine and cosine amplitudes. Positivity of $M$ and of the spring potential energy ensures that all squared angular frequencies are positive.

If instead one takes the PDF's printed $I_1=m_1{a}^2/6$ as a stipulated inertia, then $M=\mathrm{diag}(1/2,1/4,2/25)$ andThe corresponding angular frequencies are approximatelyThe same formula for $\mathbf{v}(z)$ applies, because its derivation used only the second and third rows. Here the determinant at $z=6$ is $-63/25\ne 0$, so following the printed inertia does not repair the frequency claim. The PDF therefore needs a further correction to its numerical data or asserted frequency; the intended correction cannot be inferred uniquely.