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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.

The quadratic Lagrangian gives $\partial L/\partial \dot{q}=T\dot{q}$ and $\partial L/\partial q=-Vq$. Thus the Euler-Lagrange equations areA nonzero mode vector $v$ has a common harmonic time factor and satisfies the generalized eigenvalue problem for small oscillations $Vv={\omega }^{2}Tv$; $\omega$ is its normal frequency. A zero-frequency mode instead admits a constant or linear time factor.

Measure the three coordinates as displacements from a configuration with both springs unstretched. The kinetic energy is $\frac{1}{2}{\dot{x}}_1^2+\frac{1}{2}M{\dot{x}}_2^2+\frac{1}{2}{\dot{x}}_3^2$, and the potential energy is $\frac{1}{2}(x_2-x_1{)}^2+\frac{1}{2}(x_3-x_2{)}^2$. Expansion givesFor $M>0$, direct substitution yields the three normal modes:The three vectors are independent and orthogonal in the mass inner product $v^{T}Tw$, so they exhaust the motion. Their squared mass norms are $M+2,2,2+4/M$; division by the corresponding square roots gives normalized modes. This is the unit-spring, unit-outer-mass case of the normal modes of three masses connected by two equal springs.

As $M\to \infty$, the central displacement in the symmetric oscillatory mode tends to zero and both nonzero frequencies tend to one. The two limiting shapes $(1,0,1)$ and $(1,0,-1)$ span independent motions of the outer masses against an effectively fixed centre; each unit mass has its own unit spring frequency. The translational mode remains at zero frequency. Under a bounded kinetic-energy condition its common velocity tends to zero as $M^{-1/2}$, while a static common displacement is still an equilibrium shift. One must specify that energy condition before concluding that an arbitrary freely translating heavy mass remains stationary.

The Lagrangian is $\mathcal{L}=T-\mathcal{V}$, and the Euler-Lagrange equations areAt a rest equilibrium $q=0$, ${\partial }_i\mathcal{V}(0)=0$ for all three coordinates. The Taylor expansion therefore starts asThe symmetric Hessian matrix is the matrix $\mathsf{V}$ of the quadratic potential energy. If a quadratic coefficient vanishes along a direction, higher-order terms can be the first nonconstant terms there; equilibrium by itself does not imply strict stability.

Write ${\mathsf{T}}_0=\mathsf{T}(0)$. Terms involving the variation of $\mathsf{T}(q)$ are cubic in small displacements and velocities, so the quadratic small oscillation Lagrangian isA normal mode has fixed displacement ratios, $q(t)=v\cos (\omega t+\varphi )$. Its angular frequency $\omega$ is an eigenfrequency, or normal frequency, and satisfies the generalized eigenvalue problem for small oscillationsHere ${\mathsf{T}}_0$ is positive definite for independent physical coordinates and positive masses. A stable quadratic minimum makes ${\omega }^2\ge 0$; a negative value would describe exponential instability rather than oscillation.

For the suspended bar, let ${\kappa }_0$ and ${\kappa }_1$ denote the actual incremental spring constants of the upper and each lower spring. The physical displacement of the bar is $q_1$, while the lower spring extensions change by $q_2-q_1$ and $q_3-q_1$. ConsequentlyThe linear gravitational and elastic terms cancel at the loaded equilibrium. Expanding the remaining squares gives the small-oscillation mass and stiffness matricesThusSource convention: the printed matrix is obtained by setting ${\kappa }_0=\lambda /L$ and ${\kappa }_1=\mu /l$. Although the PDF prose calls $\lambda ,\mu$ spring constants, that matrix treats them as elastic parameters whose ratios to the stated lengths are the incremental stiffnesses. If they are read literally as Hooke-law spring constants, the divisions by $L,l$ should instead be omitted. The numerical case has $L=l=1$, so this discrepancy does not affect its results.

With the numerical parameters, ${\mathsf{T}}_0=\mathrm{diag}(2,1,1)$ andUse the exchange symmetry of the lower masses to find the vertical modes of a horizontally constrained suspended bar. The antisymmetric vector $v_a=(0,1,-1{)}^T$ obeys $\mathsf{V}{v}_a={\mathsf{T}}_0{v}_a$, giving ${\omega }_a=1$. In the symmetric sector put $v=(A,B,B{)}^T$ and $s={\omega }^2$. The two independent equations areTheir determinant gives $s^2-4s+2=0$, so $s=2\pm \sqrt{2}$. Taking $A=1$, the full set isIn the lowest normal mode, all three masses move in phase, with the lower masses moving about $2.414$ times as far as the bar. In the antisymmetric normal mode, the bar stays still and the lower masses move equally in opposite directions; their changing forces on the bar cancel. In the highest normal mode, the lower masses move together against the bar, with magnitude about $0.414$ of its displacement. Any small motion is a superposition of these three modes.

For a variation $q\mapsto q+ϵ\eta$ with $\eta (t_0)=\eta (t_1)=0$, the first variation isThe endpoint term vanishes. Since the interior variation is arbitrary, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equationAlong such a trajectory,If $L$ has no explicit time dependence, the first integral is thereforeIts negative is the equivalent Beltrami form. For the usual time-independent mechanical Lagrangian $L=T-V$ with quadratic velocity dependence, this conserved quantity is the total mechanical energy $T+V$.

For a zero-natural-length spring, extension is the full distance between its endpoints, so the squared spring lengths are quadratic functions of the Cartesian coordinates. The Lagrangian isThe gravitational potential energy has positive derivative in the upward coordinate, giving downward forces. Applying the Euler-Lagrange equations to all four coordinates givesThe two spring forces are equal and opposite on the two masses for the connecting spring, as required by Newton's third law.

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

Choose positive $\varphi$ so the bob is displaced to the right relative to the cart. Its Cartesian coordinates areDifferentiation gives ${\dot{X}}^2+{\dot{Y}}^2={\dot{x}}^2+2L\cos \varphi \dot{x}\dot{\varphi }+L^2{\dot{\varphi }}^2$. Thus the exact kinetic energy and potential energy areRetain quadratic terms, using $\cos \varphi =1+O({\varphi }^2)$ in the kinetic cross term and $1-\cos \varphi ={\varphi }^2/2+O({\varphi }^4)$ in the potential. The Lagrangian isIts Euler-Lagrange equations giveThe kinetic cross term is essential: acceleration of the support couples the pendulum and cart. These equations define the normal modes of a spring-mounted pendulum cart.

For an ordinary conservative mechanical system, choose generalized coordinates $q_i$ measured from a static equilibrium point. Assume the kinetic energy is quadratic in velocities and there are no velocity-dependent forces. Expanding about equilibrium, the linear term in the potential energy vanishes. Retaining only terms of total degree two gives the small oscillation LagrangianAn irrelevant constant has been dropped. Coordinate dependence of $M$ contributes only terms of third or higher total degree. The matrices $M,K$ are symmetric, $M$ is positive definite for independent coordinates, and $K$ is positive semidefinite at a stable equilibrium. The Euler-Lagrange equations areA normal mode has a fixed mode shape $\mathbf{v}$ and one time factor: $\mathbf{q}=\mathbf{v}Q(t)$. Its angular frequency $\omega$ solves the generalized eigenvalue problem $K\mathbf{v}={\omega }^{2}M\mathbf{v}$, with $\ddot{Q}+{\omega }^{2}Q=0$. Thus $\omega$ is the normal angular frequency, and the frequency in cycles per unit time is $\omega /(2\pi )$. A zero mode satisfies $K\mathbf{v}=0$; its time dependence is $Q=A+Bt$, describing an unopposed displacement or uniform drift rather than a restoring oscillation.

Introduce the fixed endpoint values $x_0=x_{N+1}=0$. The Lagrangian isThus $T=m{I}_N$, $V_{ij}=k(2{\delta }_{ij}-{\delta }_{i,j+1}-{\delta }_{i,j-1})$, andThis is a symmetric tridiagonal matrix, reflecting that only nearest-neighbour masses are coupled.

The diagram has a connection between every pair of indices $0\le i<j\le n$, including the fixed beam $0$. Two parallel strings, each with spring constant $k/2$, together have energy $k(X_i-X_j{)}^2/2$. Thus the printed $E_{ij}$ is the energy of a connecting pair; calling it the energy of one string would introduce a factor-of-two discrepancy.

Take vertical coordinates positive upwards, with $X_0$ fixed. Each bar has gravitational potential energy $m_{i}g{X}_i$ and kinetic energy $m_i{\dot{X}}_i^2/2$. The Lagrangian isDifferentiating gives the Euler-Lagrange equationsEquivalently, with $S={\sum }_{j=1}^n{X}_j$, the spring term is $k[(n+1)X_i-S-X_0]$. Gravity changes the equilibrium but not the stiffness matrix. Subtracting a static equilibrium and writing ${\eta }_i=X_i-X_i^{\mathrm{eq}}$ gives the normal mode systemThis is the spring system underlying normal modes of an anchored complete spring graph.

Use the bead angles ${\theta }_1,{\theta }_2,{\theta }_3$ as generalized coordinates. The straight-line spring length between two beads follows from the law of cosines:The hoop is horizontal, so the gravitational potential energy is constant and can be omitted. The kinetic energy is $(m/2){\sum }_i{\dot{\theta }}_i^2$, giving the LagrangianAt ${\theta }_i^0=\Theta +2\pi (i-1)/3$, every spring has its natural length, so the potential energy is zero and its first derivatives vanish. This is an equilibrium for every constant $\Theta$. It is stable against relative displacement because each spring energy is nonnegative; the overall rotation is a neutral direction.

The generalized momentum of this Lagrangian isThe Euler-Lagrange equation therefore givesBecause $\mathcal{L}$ has no explicit time dependence, the energy form of the Beltrami identity gives a conserved quantity $E=\dot{x}\partial \mathcal{L}/\partial \dot{x}-\mathcal{L}$. HereFor a direct check of conservation of energy, differentiate:This is the rest-plus-kinetic energy of a relativistic particle, in units with light speed one, plus its potential energy.

Let $\mathbf{q}=(x,y)$ be the centre displacement and $\theta$ the small disk rotation. Put $d=l-l_0>0$. The moment of inertia of a uniform disc is $I=m{a}^2/2$, obtained from $I={\int }_0^a\rho r^2(2\pi rdr)$ and $m=\rho \pi a^2$. Thus the kinetic energy is $T=m|\dot{\mathbf{q}}{|}^2/2+m{a}^2{\dot{\theta }}^2/4$.

For one spring, let ${\mathbf{n}}_j$ point radially towards its fixed anchor and let ${\mathbf{t}}_j$ be the perpendicular unit vector. The moving attachment is $\mathbf{q}+aR(\theta ){\mathbf{n}}_j$, while the anchor is $(a+l){\mathbf{n}}_j$. Expanding their separation to second order gives the spring extension beyond natural length:Insert this in the elastic potential energy $V=(k/2){\sum }_j(s_j-l_0{)}^2$. For the three symmetric directions, ${\sum }_j{\mathbf{n}}_j={\sum }_j{\mathbf{t}}_j=0$ and ${\sum }_j({\mathbf{n}}_j\cdot \mathbf{q}{)}^2={\sum }_j({\mathbf{t}}_j\cdot \mathbf{q}{)}^2=3|\mathbf{q}{|}^2/2$. Dropping the constant equilibrium energy givesTherefore the quadratic Lagrangian isThe prestretch terms cannot be omitted: they supply transverse spring stiffness and the entire rotational restoring torque, as derived generally in small oscillations of a disk attached to prestretched radial springs.

The quadratic Lagrangian is diagonal in $x,y,\theta$, so the Euler-Lagrange equations give two pure translations and one pure rotation. Their angular frequencies areA horizontal translation has $(x,y,\theta )=(A\cos {\omega }_{x}t,0,0)$; a vertical translation has $(0,A\cos {\omega }_{y}t,0)$; and the rotational mode has $(0,0,A\cos {\omega }_{\theta }t)$. Any fixed-direction translation is another choice in the same two-dimensional degenerate normal mode space. The rotation keeps the centre fixed.

For a holonomic conservative system the Lagrangian is $L(q,\dot{q},t)=T-V$, the difference between kinetic energy and potential energy. The Euler-Lagrange equations areIf $\partial L/\partial q_j=0$, this gives $d{p}_j/dt=0$ for the generalized momentum $p_j=\partial L/\partial {\dot{q}}_j$. Such a coordinate is a cyclic coordinate.

In cylindrical coordinates the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. The holonomic constraint $z=r^2/a$ gives $\dot{z}=2r\dot{r}/a$, while gravity has potential energy $V=mgz$. ThereforeThere is no explicit $\theta$ dependence, so $\theta$ is a cyclic coordinate andThis is angular momentum about the vertical symmetry axis. At $r=0$ the polar angle is singular, but the Cartesian angular momentum $m(x\dot{y}-y\dot{x})$ remains well defined and has the same conserved value.

A coordinate absent from the Lagrangian is cyclic. Its Euler-Lagrange equation givesThus the corresponding generalized momentum $\boxed{p_1=\partial L/\partial {\dot{q}}_1}$ is conserved.

With $\mathbf{d}={\mathbf{q}}_1-{\mathbf{q}}_2$ and $r=|\mathbf{d}|$, the Lagrangian isAt configurations with $r>0$, the Euler-Lagrange equations giveThese formulas apply on the differentiable part of the potential; collision configurations require any additional smooth extension of $f(|\mathbf{d}|)$ to be specified. Summing the equations givesso the total linear momentum is conserved. The total energy isDifferentiating, the two interacting forces do work $-f^{\prime }(r)\dot{r}$, while the potential derivative is $f^{\prime }(r)\dot{r}$ and particle three contributes no force term. Hence $\dot{E}=0$. Both total momentum and total energy are conserved.

Away from a collision, let $r=q_1-q_2$ and $D=|r|>0$. The Lagrangian isDifferentiating $D^2=r\cdot r$ with respect to $x_1$ and $x_2$ gives $2D{\partial }_{x_1}D=2(x_1-x_2)$ and $2D{\partial }_{x_2}D=-2(x_1-x_2)$. ThereforeThe other coordinates give ${\nabla }_{q_1}D=r/D$ and ${\nabla }_{q_2}D=-r/D$. Substituting these into the Euler-Lagrange equations yields equal and opposite central forces:Adding the equations shows that the derivative of the total linear momentum is zero, so $P=m_1{\dot{q}}_1+m_2{\dot{q}}_2$ is constant. For the total angular momentum, the terms ${\dot{q}}_i\times {\dot{q}}_i$ vanish, givingThus both stated vectors are conserved quantities.

The total energy is also constant. Define $E=\frac{1}{2}{m}_1|{\dot{q}}_1{|}^2+\frac{1}{2}{m}_2|{\dot{q}}_2{|}^2+V(D)$. Since $\dot{D}=r\cdot ({\dot{q}}_1-{\dot{q}}_2)/D$,Hence $\boxed{E=T+V(D)\text{ is conserved}}$. The explicit cancellation establishes energy conservation for an arbitrary smooth separation-dependent potential, throughout any collision-free trajectory.

The centre of mass is at height $-a\cos \theta /2$. With gravitational potential energy zero at the pivot level,Hence the Lagrangian isAdding a constant to the potential would change the numerical energy reference but would not change the motion.

For an autonomous Lagrangian, the Euler-Lagrange equations implyIndeed, differentiating the expression cancels the acceleration terms, leaving ${\sum }_q\dot{q}(d{L}_{\dot{q}}/dt-L_q)-L_t$, which vanishes. Here $T$ is quadratic in the velocities, so $\sum \dot{q}{L}_{\dot{q}}=2T$. Thus the conserved energy isThe azimuth $\varphi$ is a cyclic coordinate. Its conjugate momentum is constant:Therefore $\boxed{{\sin }^2\theta \dot{\varphi }=h}$ is constant as well. Physically, $p_{\varphi }$ is the vertical component of angular momentum.