= Solution
Substitution of $\mu=2$ into the <normal modes of three masses connected by two equal springs> gives
$$
\boxed{\Omega=\sqrt{\frac km},\qquad\omega_0=0,\quad\omega_1=\Omega,\quad\omega_2=\sqrt2\,\Omega.}
$$
The corresponding shapes are $v_0=(1,1,1)^T$, $v_1=(1,0,-1)^T$, $v_2=(1,-1,1)^T$. For example, $Kv_2=k(2,-4,2)^T=2(k/m)Mv_2$, verifying the third <eigenvalue> directly.
Their relevant <orthogonality> uses the kinetic matrix $M=m\operatorname{diag}(1,2,1)$:
$$
\boxed{v_i^TMv_j=0\quad(i\ne j),\qquad\begin{pmatrix}v_0&v_1&v_2\end{pmatrix}^{\!T}M\begin{pmatrix}v_0&v_1&v_2\end{pmatrix}=m\operatorname{diag}(4,2,4).}
$$
This is <mass-matrix orthogonality of normal modes>. It also follows by subtracting $v_i^TKv_j=\omega_j^2v_i^TMv_j$ from its transpose and using the symmetry of $K,M$. Ordinary <Euclidean inner products> need not vanish: in fact $v_0^Tv_2=1$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2-2017-ib-normal-modes.png]
{title=The three <normal modes> for a free spring chain with middle mass $2m$; amplitudes are relative and the zero mode is translation}
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