Lagrangian mechanics

Lagrangian mechanics derives motion from the stationary action of $L=T-V$ through the Euler--Lagrange equations.

A massless fixed-axis pulley carries a rim point mass $M$ at radius $R$. A mechanical string gripping its rim without slipping passes over a fixed guide pulley to a hanging mass $m$. Choose the angle $\theta$ from downward vertical so the hanging mass descends by $R\theta$. The Lagrangian is $(M+m)R^2{\dot{\theta }}^2/2+MgR\cos \theta +mgR\theta$. The stable equilibrium satisfies $\sin {\theta }_e=m/M$ with $\cos {\theta }_e>0$, and exists for $0<m<M$. Its small-amplitude angular frequency is $[Mg\cos {\theta }_e/((M+m)R){]}^{1/2}$.

An inextensible uniform chain of length L and linear mass density $\rho$ links a mass m sliding on a frictionless table to a hanging mass M. If the initial hanging chain length is zero and x is its subsequent hanging length, all moving material has speed $\dot{x}$. The kinetic energy is $(M+m+\rho L){\dot{x}}^2/2$, while the gravitational potential energy is $-Mgx-\rho g{x}^2/2$. The Euler-Lagrange equation gives the displayed equation until a mass hits the floor or reaches the edge. Increasing hanging chain weight produces a hyperbolic cosine displacement from rest, rather than uniform acceleration.

Attach the table-end mass to a linear spring of spring constant k. Adding spring potential energy $k(e_0+x{)}^2/2$ gives restoring curvature $k-\rho g$ about equilibrium. The positive gravitational feedback from increasing hanging chain length subtracts from the spring stiffness. Stable oscillations require $k>\rho g$; equality is neutral in this ideal linear model, while $k<\rho g$ gives exponential departure. The frequency is independent of the preload, although the equilibrium position is not.

A reduction of Lagrangian mechanics that fixes conserved momenta of cyclic coordinates while keeping an Euler-Lagrange description for the remaining coordinates. It is named for Edward Routh. The Routhian applies a partial Legendre transform to the eliminated velocities.

Use the convention $R=L-{\sum }_i{p}_i{\dot{q}}_i$ for the eliminated cyclic coordinates, expressing their velocities in terms of their fixed conjugate momenta. Terms involving derivatives of those eliminated velocities cancel because $\partial L/\partial {\dot{q}}_i=p_i$. Thus the remaining coordinates obey the usual Euler-Lagrange equations for R. For a particle of fixed angular momentum $\ell$ in a central potential, this gives $R=m{\dot{r}}^2/2-V(r)-{\ell }^2/(2m{r}^2)$, rather than the expression obtained by direct substitution into L alone. An opposite overall sign convention is also used, with correspondingly signed mixed Hamiltonian equations.

For a Lagrangian $L(q,\dot{q},t)$, its energy function satisfies $d\mathcal{E}/dt=-\partial L/\partial t$ along the Euler-Lagrange equations. This follows by differentiating and cancelling the ${\ddot{q}}_i\partial L/\partial {\dot{q}}_i$ terms. It is conserved when $L$ has no explicit time dependence. For imposed moving constraints it need not coincide with inertial mechanical energy; a Jacobi integral is an important example.

For a uniform slender ladder of length $2l$ sliding while touching a smooth wall and floor, use its angle $\theta$ from the vertical. Its centre of mass is $(l\sin \theta ,l\cos \theta )$ and its central moment of inertia is $m{l}^2/3$. Hence the Lagrangian is $L=2m{l}^2{\dot{\theta }}^2/3-mgl\cos \theta$, and its Euler-Lagrange equation is $\ddot{\theta }=3g\sin \theta /(4l)$. The contacts are physically possible only while their normal forces are nonnegative. If released from rest at ${\theta }_0$, energy conservation gives ${\dot{\theta }}^2=3g(\cos {\theta }_0-\cos \theta )/(2l)$, and the wall force becomes $3mg\sin \theta (3\cos \theta -2\cos {\theta }_0)/4$. Contact is lost at $\cos \theta =2\cos {\theta }_0/3$, after which the constrained equation no longer describes the motion.

A point mass attached to a massless straight spring can change its radial distance $r$ from a fixed support while swinging in a vertical plane. Its lower equilibrium length is $r_0=l+mg/k$. Linearizing the Euler-Lagrange equations there gives independent radial and angular normal modes with squared angular frequencies $k/m$ and $g/r_0$. The formal upper stationary state has $r=l-mg/k$ and exists in positive-length coordinates only when $l>mg/k$; its angular mode is unstable.

A tabletop mass $m$ joined through a hole by a massless inextensible string of length $l$ to a hanging mass $M$ has radial inertia $M+m$ and conserved angular momentum $J=m{r}^2\dot{\theta }$. The radial energy is $\frac{1}{2}(M+m){\dot{r}}^2+U_J(r)$, with $0<r<l$. Its effective potential diverges at the hole for $J\ne 0$ and has a unique minimum at $r_0^3=J^2/(mMg)$. Since $U_J^{\prime \prime }(r_0)=3Mg/r_0$, small radial oscillations at fixed $J$ have squared angular frequency $3Mg/[(M+m)r_0]$.

An Atwood machine couples two hanging masses by an inextensible mechanical string over a pulley. With ideal massless hardware, Newton's second law gives the downward acceleration of the heavier mass as $a=g(m_2-m_1)/(m_1+m_2)$. To prove it, write $m_{2}g-T=m_{2}a$ and $T-m_{1}g=m_{1}a$, then add. The apparatus associated with George Atwood turns a small difference of weights into slow, measurable acceleration.

Two initially equal buckets of total mass $M_0$ share an ideal Atwood machine. If one loses sand at constant rate $\alpha$ and the sand leaves with zero initial velocity relative to that bucket, the leakage supplies no rocket thrust. The remaining masses satisfy $(M+M_0)\dot{v}=g(M_0-M)$ for upward velocity of the leaking side. Using $\dot{M}=-\alpha$ and integrating from rest gives $v=(g/\alpha )[M-M_0-2M_0\log ((M+M_0)/(2M_0))]$.

If the initial sand fraction is $ϵ$, emptying occurs at $M=(1-ϵ)M_0$. This gives the displayed speed, which tends to zero for $ϵ\to 0$ and to $(g{M}_0/\alpha )(2\log 2-1)$ for $ϵ\to 1^{-}$. The latter is a finite limiting result, not a physical claim about an exactly massless empty bucket. The derivation is an application of a variable-mass system, rather than of $d(Mv)/dt=F$ for a closed body.

A compound Atwood machine replaces one hanging mass by another pulley carrying two masses. Let the upper mechanical string carry mass $M$ opposite the moving pulley, and let the lower mechanical string carry masses $a,b$ at upward heights $x,y$. The length constraints on the mechanical strings give the height of $M$ as $-(x+y)/2$ up to a constant. Thus its Lagrangian isThis follows by adding the three kinetic energies and subtracting their gravitational potential energies. Constant coordinate shifts annihilating the linear potential energy give conserved generalized momentum components by Noether's theorem.

For a mechanical system whose imposed motion is uniform rotation with angular velocity $\omega$ about a fixed axis, a conserved rotating-frame energy can exist even when the inertial mechanical energy $E$ changes. With angular momentum $L_z$ about that axis, it is $J=E-\omega L_z$. For a bead constrained to rotate with $\dot{\theta }=\omega$, this becomesFor $z=r^2/(2a)$ in uniform gravity, division by mass yieldsThis is the first integral that controls the bead's radial motion. Different normalizations and signs are used for the related Jacobi constant in celestial mechanics.

For a pendulum whose support slides without horizontal force, the support coordinate is cyclic and total horizontal momentum is conserved. At zero horizontal momentum, eliminating support velocity leaves the displayed angle-dependent effective inertia. The energy is $M_{\mathrm{eff}}{\dot{\varphi }}^2/2-m_{2}gl\cos \varphi$. Fixing the momentum before evaluating motion is distinct from fixing the support position.

Zero horizontal momentum fixes the horizontal centre of mass. The bob coordinates become $X=X_c+l{m}_1\sin \varphi /(m_1+m_2)$ and $Y=-l\cos \varphi$. Eliminating the angle gives an ellipse; an ordinary oscillation traverses only its permitted arc. The support counter-moves, so the bob's path differs from the circle about a fixed pivot.

A Lagrangian is degenerate when its Hessian with respect to velocities is singular. The momentum–velocity relation then cannot be inverted for every velocity. Some Euler–Lagrange equations can become algebraic constraints rather than independent second-order equations. A massless limit can produce this behaviour; disappearance of an equation at quadratic order does not establish an unconstrained mode of the exact nonlinear model.

A uniformly charged circular ring hasHere $k$ is the spring constant for extension of its full circumference and the symmetric gauge is used. The canonical angular momentum $J=m{r}^2\dot{\varphi }+qB{r}^2/2$ is conserved. Eliminating the angle gives $m\ddot{r}=J^2/(m{r}^3)-q^2{B}^{2}r/(4m)-4{\pi }^{2}k(r-r_0)$. This model neglects electric self-interaction and noncircular deformation.

For negligible natural radius and release from rest at $r_1$, $J=qB{r}_1^2/2$. The effective radial potential is a sum of $r^{-2}$ and $r^2$ terms, whose two turning radii have squared product $J^2/(m^2{\Omega }^2)$. HenceThese formulas assume $qB\ne 0$. At exactly zero field the ideal ring collapses to zero radius instead of making a regular positive-radius turning point.

With $J=0$, define ${\omega }_s^2=4{\pi }^{2}k/m$ and ${\omega }_c=qB/m$. Then $r_e={\omega }_s^2{r}_0/({\omega }_s^2+{\omega }_c^2/4)$, the radial frequency is $\Omega =\sqrt{{\omega }_s^2+{\omega }_c^2/4}$, and release at $r_0$ gives $r=r_e+(r_0-r_e)\cos \Omega t$ with $\dot{\varphi }=-{\omega }_c/2$. The physical solution requires $r>0$. If $2r_e<r_0$, the formal oscillation reaches zero and cannot be continued as a negative ring radius without a new collapse model.

Although the energy in velocities is $m({\dot{r}}^2+r^2{\dot{\varphi }}^2)/2+2{\pi }^{2}k(r-r_0{)}^2$, the Hamiltonian in canonical variables isThe field therefore enters Hamilton's equations through the relation between canonical and mechanical momentum, even though a magnetic field does no work.

A constraint that depends explicitly on time can exert a normal force that does actual work, even when its force does no work on an allowed instantaneous virtual displacement. The surface motion gives the constrained particle a normal component of actual velocity.

A point mass $M$ attached to a massless rotating hoop of radius $R$ is connected by a taut string wrapping around the hoop to a vertically suspended mass $m$. Measure $\theta$ from the downward radius toward the string's tangency side. The wrapped length decreases by $R\theta$, so the suspended mass descends by the same amount. Therefore $T=(M+m)R^2{\dot{\theta }}^2/2$ and $V=-MgR\cos \theta -mgR\theta$ up to a constant. The Euler-Lagrange equation gives the displayed equation. For $0<m<M$, the stable equilibrium has $\sin {\theta }_0=m/M$, and the small-oscillation angular frequency is $\omega =\sqrt{g/R}[(M-m)/(M+m){]}^{1/4}$. When $M\gg m$, the system approaches a simple pendulum of length $R$.

A generalized coordinate is one of a set of independent parameters specifying the configuration of a mechanical system subject to its constraints. An angle locates a bead constrained to a fixed circular hoop. Expressing kinetic energy and potential energy in these coordinates gives the Lagrangian and its Euler-Lagrange equations.

The time derivative of a generalized coordinate. It is an independent velocity argument of a mechanical Lagrangian, but need not have the units or direction of a Cartesian velocity: for an angular coordinate it is an angular velocity. A regular Legendre transform replaces these variables with the corresponding canonical momenta.

A regular Lagrangian $L:TQ\to \mathbb{R}$ defines a fibre derivative $\mathbb{F}L:TQ\to T^{\ast }Q$ and the Cartan one-form ${\theta }_L=(\mathbb{F}L{)}^{\ast }\theta$. Its two-form ${\omega }_L=-d{\theta }_L$ is nondegenerate when the velocity Hessian is invertible. The energy is $E_L=v^i{L}_{v^i}-L$. The equation ${\iota }_{{\Gamma }_L}{\omega }_L=d{E}_L$ determines the Euler–Lagrange vector field. A hyperregular Lagrangian makes the fibre derivative a global diffeomorphism and transports this dynamics to Hamiltonian phase space.

A potential $V(r,\varphi ,z)=f(r,\varphi -kz)$ is invariant under the screw transformation $(\varphi ,z)\mapsto (\varphi +ks,z+s)$. The Noether conserved quantity for a mechanical point symmetry isa linear combination of axial linear momentum and axial angular momentum.

A generalized coordinate absent from the Lagrangian is cyclic. Its conjugate generalized momentum is conserved by the Euler-Lagrange equations.

The action of a path is the time integral of its Lagrangian. The principle of stationary action requires its first variation to vanish for every admissible fixed-endpoint variation.

The principle of stationary action says that the physical path between fixed endpoints is a stationary point of the action.

Replacing a Lagrangian byadds only the endpoint term $F(q(t_2),t_2)-F(q(t_1),t_1)$ to the action. Fixed-endpoint variations therefore give the same Euler--Lagrange equations.

For electromagnetic scalar and vector potentials $\varphi$ and $A$,Withthe Euler--Lagrange equations give the Lorentz force $m\ddot{r}=q(E+\dot{r}\times B)$, while the canonical momentum is $p=m\dot{r}+qA$.

The Legendre transform of the charged-particle electromagnetic Lagrangian is

Under $\varphi \mapsto \varphi -{\partial }_{t}f$ and $A\mapsto A+\nabla f$, the charged-particle Lagrangian changes byIt therefore has unchanged equations of motion by total-time-derivative invariance.

For a constant uniform magnetic field $B$, choose the symmetric gauge $A=\frac{1}{2}B\times r$. The coordinate parallel to $B$ is cyclic, so its canonical momentum is conserved. Since $A\cdot B=0$, this component is also the mechanical momentum $m\dot{r}\cdot \hat{B}$.

The generalized, or conjugate, momentum associated with a coordinate $q_i$ is

In Lagrangian mechanics, the coordinate variable describes the configuration; its conjugate momentum need not equal the Cartesian mechanical momentum.

A coordinate $q_i$ is ignorable, or cyclic, when $\partial L/\partial q_i=0$. The Euler--Lagrange equation then gives ${\dot{p}}_i=0$, so its generalized momentum is conserved.

If a Lagrangian is independent of a coordinate, the conjugate momentum to that coordinate is conserved by the Euler--Lagrange equations.

If $L=|\dot{q}{|}^2/2-V(q)$ and $V$ depends only on $|q{|}^2$ and $(q\cdot n{)}^2$, then rotations about the axis $n$ leave $L$ invariant. The quantity supplied by Noether theorem is the axial angular momentumIf $L$ also has no explicit time dependence, the energy $|\dot{q}{|}^2/2+V(q)$ is conserved independently.

A double pendulum consists of two pendula coupled because the second pivot moves with the first mass. It has two degrees of freedom: even when the upper pivot is fixed, moving the upper arm carries the lower pivot along. This makes the double pendulum a compact example of how mechanical constraints couple motion, and how Lagrangian mechanics converts geometry into dynamics.

Use two equal point masses $m>0$ and two massless rigid rods of equal length $\ell >0$, with joints that allow free rotation in one vertical plane. The upper pivot is fixed; gravity is uniform, with gravitational acceleration $g>0$. Neglect joint friction and drag force. Both angles ${\theta }_1,{\theta }_2$ are measured from the downward vertical, positive toward the right, in radians. In particular, ${\theta }_2$ is an absolute angle, not the bend between the rods: that bend is ${\theta }_2-{\theta }_1$.

The double pendulum Lagrangian below gives the exact dynamics of this ideal model. Near the downward stable equilibrium, normal modes of a double pendulum replace the coupled motion with two independent harmonic oscillators. Their fixed amplitude ratios explain which releases make the rods swing together and which make them swing in opposite directions.

With equal rod lengths $\ell$ and absolute downward-vertical angles $\theta ,\varphi$, the endpoint separation is $s^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. A zero-rest-length spring between the fixed support and the final mass contributes $-\kappa {\ell }^2[1+\cos (\theta -\varphi )]$ to the Lagrangian. It favours bending away from maximum extension and can reduce the gravitational bending stiffness.

Choose Cartesian coordinates with origin at the upper pivot, $x$ to the right and $y$ upward. The two position vectors areThese expressions satisfy the rod holonomic constraints $|{\mathbf{r}}_1|=\ell$ and $|{\mathbf{r}}_2-{\mathbf{r}}_1|=\ell$ automatically. This is the practical advantage of using the two angles as coordinates: the rod tensions need not be unknowns in the motion calculation.^[ref]

Differentiate these position vectors to obtain the velocities. The dot product between their two rod contributions uses $\cos ({\theta }_1-{\theta }_2)=\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2$, givingAdding the two kinetic energies and measuring gravitational potential energy from the downward equilibrium givesThe coefficient $2$ has a geometric meaning: the upper rod raises both masses, and its motion contributes to both kinetic energies. The coupling in $T$ is present because the lower mass has a moving pivot. The zero chosen for $V$ changes $L$ by a constant and leaves the Euler-Lagrange equations unchanged.^[ref]

The principle of stationary action applies to $S=\int Ldt$. Its fixed-endpoint first variation, after integration by parts, isIndependent allowed variations therefore give one Euler-Lagrange equation for each angle. Stationary action alone does not assert a minimum.^[ref]

Write $\Delta ={\theta }_1-{\theta }_2$. Applying the Euler-Lagrange equations to the double pendulum Lagrangian, then dividing by $m{\ell }^2$, givesFor example, the upper generalized momentum is $\partial L/\partial {\dot{\theta }}_1=m{\ell }^2(2{\dot{\theta }}_1+\cos \Delta {\dot{\theta }}_2)$. Its time derivative contains $-m{\ell }^2\sin \Delta ({\dot{\theta }}_1-{\dot{\theta }}_2){\dot{\theta }}_2$. Subtracting $\partial L/\partial {\theta }_1=-m{\ell }^2\sin \Delta {\dot{\theta }}_1{\dot{\theta }}_2-2mg\ell \sin {\theta }_1$ cancels the mixed product and leaves the positive $\sin \Delta {\dot{\theta }}_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.

The matrix multiplying the two angular accelerations isIt is invertible for every configuration, so specified angles and angular velocities determine both angular accelerations. For the same reason $T>0$ whenever the angular velocities are not both zero. The mass cancels from the motion equations, while the timescale is set by $\sqrt{\ell /g}$.

As an exact constructed release, put ${\theta }_1(0)=\alpha$, ${\theta }_2(0)=0$, and both angular velocities zero. Solving the two equations at release yieldsFor $0<\alpha <\pi /2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's acceleration makes this possible even though the lower rod starts vertical. The small-angle approximation gives ${\ddot{\theta }}_1(0)\simeq -2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)\simeq 2g\alpha /\ell$.

The conserved mechanical energy is $E=T+V$, because the Lagrangian has no explicit time dependence. Directly, $dE/dt$ is $m{\ell }^2$ times the first equation multiplied by ${\dot{\theta }}_1$ plus the second multiplied by ${\dot{\theta }}_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a trajectory.

The University of Toronto's institutional demonstration “Chaotic Behaviour (Double Pendulum)”, activity Me-020, uses a metal stand with four leveling feet and a pendulum arm carrying a white dot. Preparation includes leveling the stand and correcting any wobble. The instructions call for holding the arm vertically upward and releasing it; the readout is the visible motion of the dot. A further comparison inserts a metal dowel in a groove on the release arm and repeats the release at the same location. Keeping the stand steady and repeating the release position provide practical controls. The page supplies no numerical angle trace or calibrated timing, and does not explain the dowel's mechanical effect. It therefore documents a qualitative release demonstration, without quantitatively testing the equal-point mass model. Its extended arms also have moments of inertia absent from massless rods. Irregular visible motion alone does not establish sensitive dependence on initial conditions.^[ref]

A separate comparison with normal modes of a double pendulum would require small downward releases and measurement of both angles over time. The actual arm masses, centres of mass and moments of inertia would first have to enter the Lagrangian. The upward release in this demonstration explores a different regime from small oscillations about the downward stable equilibrium.

For point masses with generalized coordinates, kinetic energy follows by differentiating their Cartesian position vectors and summing $m_i|{\dot{r}}_i{|}^2/2$.

Near a stable equilibrium, retaining quadratic Lagrangian terms gives a linear system $\mathsf{M}\ddot{z}+\mathsf{K}z=0$.

The resulting linearized motion decomposes into normal modes near the stable equilibrium.

Two frictionless beads of positive masses $m_1,m_2$ on a circular hoop are joined by two identical springs of spring constant $k$, constrained to follow the two arcs. At their diametrically opposite equilibrium, each spring has natural length equal to half the circumference. For signed arc length displacements in the same rotational sense, the potential energy is $U=k(x_1-x_2{)}^2$. The normal mode of common rotation has mode shape $(1,1)$ and zero angular frequency, while the other normal mode has mode shape $(1,-m_1/m_2)$ and the positive angular frequency in the title.

Adding the equations of motion shows that $X=(m_1{x}_1+m_2{x}_2)/(m_1+m_2)$ satisfies $\ddot{X}=0$. Subtracting them gives $\ddot{\delta }+{\omega }_1^2\delta =0$ for $\delta =x_1-x_2$. Thus the complete zero mode is $X=C+Dt$, including uniform rotation, and the relative motion is a harmonic oscillator. In the limit $m_2/m_1\to \infty$ at fixed $m_1,k$, the oscillatory motion of the heavier bead vanishes for fixed lighter-bead amplitude, and ${\omega }_1\to \sqrt{2k/m_1}$. A common drift still survives if its initial velocity is held fixed.

Four equal beads of mass $m$ are connected cyclically by identical springs of stiffness $k$ following arcs of a frictionless circular guide. In signed arc displacements $q_j$, the quadratic Lagrangian is $L=(m/2)\sum {\dot{q}}_j^2-(k/2)\sum (q_{j+1}-q_j{)}^2$. The stiffness is $k$ times the four-cycle graph Laplacian. Its eigenvectors can be chosen as $(1,1,1,1)/2$, $(1,0,-1,0)/\sqrt{2}$, $(0,1,0,-1)/\sqrt{2}$, and $(1,-1,1,-1)/2$, with squared frequencies in the title. Their dot products with $q$ are decoupled normal mode coordinates. The zero mode is a rigid rotation; the two middle modes span a degenerate eigenspace; the highest mode alternates adjacent arc displacements. Changing the guide radius multiplies angular inertia and angular spring energy by the same factor, leaving these frequencies unchanged.

Near a stable equilibrium $x_0$ with $V^{\prime }(x_0)=0$ and $V^{\prime \prime }(x_0)>0$, the harmonic approximation truncates the Taylor series of the potential energy at quadratic order. With constant mass $m$, Newton's second law then gives a harmonic oscillator with angular frequency $\omega =\sqrt{V^{\prime \prime }(x_0)/m}$. The approximation requires displacements small enough that the omitted terms are negligible; it does not describe a flat minimum where $V^{\prime \prime }(x_0)=0$.

If the quadratic kinetic energy has a singular coefficient matrix, some generalized coordinates have no independent linearized inertia. The small oscillation equations can then contain algebraic constraints, and the number of finite normal mode frequencies can be smaller than the number of coordinates. One should use the polynomial $\det (\mathsf{K}-{\omega }^2\mathsf{M})$ without dividing by $\det \mathsf{M}$. Giving a nominally massless component a small mass can restore a high-frequency mode whose frequency diverges as that mass tends to zero.

A cart of mass $m$ moves horizontally on a spring of stiffness $k$, carrying a simple pendulum of bob mass $M$ and length $L$. Let $x$ be cart displacement and $\varphi$ the small angle from downward vertical. The quadratic Lagrangian isThe kinetic cross term comes from the bob's horizontal velocity $\dot{x}+L\dot{\varphi }$ to first order. The generalized eigenvalue problem has mass matrix $\left(\begin{array}{cc}m+M& ML\\ ML& M{L}^2\end{array}\right)$ and restoring matrix $\mathrm{diag}(k,MgL)$. Their characteristic determinant gives the displayed polynomial in ${\omega }^2$. For $m=M=L=g=1,k=2$, the squared angular frequencies are $2\mp \sqrt{2}$, with mode shapes $(x,\varphi )=(1,\pm \sqrt{2})$. The bob and cart move in phase in the lower normal mode and in opposite phases in the higher normal mode.

For a bob of mass $m$ on a massless rod of length $\ell$, let a massless support slide horizontally against a spring of stiffness $k$. With support displacement $s=\ell \eta$ and angle $\theta$ from downward vertical, the quadratic Lagrangian, divided by $m{\ell }^2$, is $(\dot{\eta }+\dot{\theta }{)}^2/2-(k/m){\eta }^2/2-(g/\ell ){\theta }^2/2$. Subtracting its two Euler-Lagrange equations gives $\eta =mg\theta /(k\ell )$. There is one finite normal mode, with the displayed frequency. A stiff spring fixes the support and recovers the simple pendulum frequency. The missing second finite frequency reflects a singular kinetic matrix in small oscillations, not an omitted oscillation branch.

Consider three beads of masses $m,m,M$ on a horizontal hoop, connected in cyclic order by identical springs constrained to follow the arcs. Their common natural length affects only a constant in the potential energy, since the sum of the three arc lengths is fixed. About the equally spaced configuration, angular displacements $q_i$ haveThe generalized eigenvalue problem has the rigid rotation $(1,1,1)$ as a zero mode. Exchanging the equal masses separates an antisymmetric normal mode $(1,-1,0)$, with squared angular frequency $3k/m$, from a symmetric normal mode $(1,1,-2m/M)$, with squared angular frequency $k/m+2k/M$. The last mode shape has zero total angular momentum. The hoop radius cancels from both the kinetic energy and the restoring quadratic form. These formulas use arc-constrained springs; straight chord springs have a different potential energy.

For point masses on concentric horizontal rings in parallel planes, set $I_1=m{a}^2$, $I_2=M{A}^2$, and $\delta =\theta -\phi$. A spring of zero natural length gives $V=\text{constant}-kaA\cos \delta$. The inertia-weighted angle $q=(I_1\theta +I_2\phi )/(I_1+I_2)$ separates the Lagrangian into a free common rotor with inertia $I_1+I_2$ and a relative rotor with inertia $I_r=I_1{I}_2/(I_1+I_2)$. Static equilibria have ${\delta }_0=0$ or $\pi$. Their normal modes are the common displacement $(1,1)$ with zero frequency and the relative displacement $(I_2,-I_1)$ with the displayed squared frequency. The aligned family has a restoring relative mode; the antipodal family has an exponentially growing one. The zero mode is required by rotational symmetry. Small nonzero total angular momentum produces common drift, so stability of the aligned family modulo rotations should be distinguished from Lyapunov stability of one fixed equilibrium point.

A uniform rigid body disk of mass $m$ and radius $a$ moves in a frictionless plane. Let $N\ge 3$ identical springs attach equally spaced rim points to fixed anchors in their radial directions. The equilibrium spring length is $l$, natural length $l_0<l$, and spring constant $k$. Write the small centre displacement as $\mathbf{q}$ and the rotation as $\theta$. For a spring with radial and tangential unit vectors $\mathbf{n},\mathbf{t}$, its length isThis follows by a second-order Taylor series for the distance between $\mathbf{q}+aR(\theta )\mathbf{n}$ and $(a+l)\mathbf{n}$. In the elastic potential energy $k(s-l_0{)}^2/2$, the equilibrium tension $k\tau$ multiplies the second-order change of length. Thus it contributes to the linearized stiffness even though the displacement is small.

The equally spaced directions satisfy $\sum \mathbf{n}=\sum \mathbf{t}=0$ and $\sum \mathbf{n}{\mathbf{n}}^T=\sum \mathbf{t}{\mathbf{t}}^T=(N/2)I$. Summing the spring energies givesThe kinetic energy is $m|\dot{\mathbf{q}}{|}^2/2+m{a}^2{\dot{\theta }}^2/4$, using the moment of inertia of a uniform disc. The Euler-Lagrange equations give the displayed squared angular frequencies. The translational normal modes are degenerate; rotation is independent. At zero prestretch the rotational frequency vanishes at quadratic order, and finite-angle spring stretching first contributes at higher order.

At an equilibrium, a positive-definite Hessian matrix of the potential gives stable small oscillations. If a continuous symmetry supplies zero modes, positive definiteness on the directions transverse to the symmetry gives stability of the relative configuration.

Angular frequency measures phase advance per unit time. For $x(t)=A\cos (\omega t+\phi )$ it is $\omega =2\pi /T$, where $T$ is the period.