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For a conservative system near a stable equilibrium, the quadratic Lagrangian has the form $L=\frac{1}{2}{\dot{q}}^{T}M\dot{q}-\frac{1}{2}{q}^{T}Kq$, with symmetric mass matrix $M$ and stiffness matrix $K$. A normal mode $q=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem for small oscillations $(K-{\omega }^{2}M)v=0$, so the frequency equation is $\det (K-{\omega }^{2}M)=0$.

To quadratic order, vertical velocities contribute only fourth-order corrections to the kinetic energy, leaving $T=({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)/2$. A string of length $\ell$ with small horizontal difference $d$ shortens its vertical projection by $d^2/(2\ell )$. The upper segment carries all three masses, the middle segment two, and the lower segment one. Hence their increases in gravitational potential energy sum toSubtracting $V$ from $T$ gives the stated approximate Lagrangian. Here $M=I$ andFor a mode with middle displacement zero, write $v=(X,0,Z{)}^T$. The middle row requires $X/b+Z/c=0$, so $Z=-cX/b$. A nonzero mode then has $X,Z\ne 0$, and the outer rows require ${\omega }^2/g=a^{-1}+b^{-1}=c^{-1}$. Therefore the necessary and sufficient condition isWhen it holds, the mode shape is $(1,0,-c/b)$ with $\omega =\sqrt{g/c}$. This is the force-cancellation mechanism of the stationary middle mass in a three-mass spring chain, applied to the equivalent quadratic stiffnesses here.

With downward position as the vertical coordinate, the full potential energy isThe equilibrium tensions are $2k(a-\ell )=6mg$, $k(b-\ell )=3mg$ and $k(c-\ell )=mg$. On expanding, these relations cancel every term linear in the displacements, leaving $V=k{x}^2+k(x-y{)}^2/2+k(y-z{)}^2/2+V_0$. Newton's second law therefore givesFor a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, set $s=m{\omega }^2/k$. The generalized eigenvalue problem isIts determinant is $2(1-s)[3(1-s{)}^2-2]$, so the angular frequencies and mode directions areThe square roots of these positive values give the requested frequencies. Only the lowest mode has all components of the same sign, so all three particles oscillate in phase at the lowest frequency. The mode shapes are shown below.

For release from rest with $Y$ remaining at equilibrium, chooseThis excites only the middle-frequency mode, so $y(t)=0$ for all time. It is a stationary middle mass in a three-mass spring chain: the equal and opposite adjacent spring forces cancel on $Y$.