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Releasing the central bob of three identical spring-coupled simple pendulums from displacement $d$ excites only the symmetric normal modes. In displacement ratios, $(0,d,0)=(d/3)(1,1,1)-(d/3)(1,-2,1)$. Thus the outer coordinates are $(d/3)(\cos {\omega }_{1}t-\cos {\omega }_{3}t)$ and the central coordinate is $(d/3)(\cos {\omega }_{1}t+2\cos {\omega }_{3}t)$, where ${\omega }_1^2=g/l$ and ${\omega }_3^2=g/l+3k/m$. The antisymmetric mode is absent by reflection symmetry. The central modulation envelope is $|d|\sqrt{5+4\cos [({\omega }_3-{\omega }_1)t]}/3$, so it never vanishes completely. Slow beats require a small frequency difference; the decomposition itself is exact in the linear model.

A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

Dimensional analysis checks the physical dimensions of quantities and identifies combinations unchanged by the choice of units. It catches impossible equations and reveals scaling before a full dynamical calculation. Adding two quantities requires the same physical dimension; replacing metres by centimetres must change numerical values consistently while preserving physical predictions.

For the ideal simple pendulum, suppose its oscillation period $T$ depends only on mass $m$, length $\ell$, gravitational acceleration $g$, and dimensionless angular amplitude ${\theta }_0$. Write $T=m^a{\ell }^b{g}^{c}F({\theta }_0)$. Since $[m]=M$, $[\ell ]=L$, and $[g]=L{T}^{-2}$, matching physical dimensions gives $a=0$, $b+c=0$, and $-2c=1$. ConsequentlyThe angular amplitude is dimensionless when expressed in radians. Dimensional analysis establishes the length/gravity scaling and the absence of a mass factor under these assumptions, but cannot determine $F$. The harmonic oscillator equation gives $F\to 2\pi$ at small amplitude; the full simple pendulum supplies its finite-amplitude correction. A model with drag or a finite massive support has additional parameters, so the same dimensional argument does not automatically apply.

The result is useful for comparison: at fixed angular amplitude and $g$, lengths in the ratio $1:4$ give periods in the ratio $1:2$. The ratio $T\sqrt{g/\ell }$ should be the same for both in this ideal model. A failure to collapse onto the same value points to measurement error or an omitted parameter. Correct physical dimensions are a necessary consistency check, while the coefficient and the adequacy of the model require more information.

Two identical planar simple pendulums of length $l$ and bob mass $m$, with pivots separated horizontally by $r>0$, are coupled by a straight spring of natural length $r$ and spring constant $k$. In the ideal model without contact restrictions, their configuration space is the torus $S^1\times S^1$. Measure both angles from the downward vertical towards the same horizontal direction. Their bob separation isThe exact Lagrangian is the expression in the title. To derive it, use bob positions $(l\sin {\theta }_1,-l\cos {\theta }_1)$ and $(r+l\sin {\theta }_2,-l\cos {\theta }_2)$. Kinetic energy from Cartesian position vectors gives $T=m{l}^2({\dot{\theta }}_1^2+{\dot{\theta }}_2^2)/2$. Gravitational potential energy is $-mgl(\cos {\theta }_1+\cos {\theta }_2)$ and spring potential energy is $k(d-r{)}^2/2$; subtracting these from $T$ proves the formula. Near the downward equilibrium, $d-r=l({\theta }_2-{\theta }_1)+O(|\mathbf{\theta }{|}^3)$ for fixed $r$, giving the familiar normal modes of two spring-coupled equal pendulums.

A Foucault pendulum is a long simple pendulum whose suspension allows the bob to swing in both independent horizontal directions. Its slowly changing swing direction relative to the building reveals Earth's rotation. A hinge restricting the bob to one vertical plane would prevent this measurement: the support would impose the very direction we want the pendulum to select freely. Léon Foucault demonstrated the effect in 1851.^[ref]

The bob's back-and-forth oscillation and the swing direction's precession are different motions. The bob traverses its arc in seconds, while successive arcs change direction over hours. In a conical pendulum the bob itself circulates around the vertical; a nearly planar Foucault pendulum instead reverses along an almost straight horizontal trace, whose orientation changes slowly. The measured quantity is that orientation against a fixed floor or wall reference, rather than the bob's instantaneous azimuth.

For small angles about the downward vertical and prescribed pivot position $x=A\cos \omega t$, a simple pendulum obeys the displayed forced-oscillator equation. Off resonance, its harmonic particular solution is $[A{\omega }^2/(l({\omega }_0^2-{\omega }^2))]\cos \omega t$, with ${\omega }_0^2=g/l$. At exact resonance, a pendulum initially vertical with zero angular velocity has $\theta =(A{\omega }_0/(2l))t\sin {\omega }_{0}t$ in the undamped linear model. The growing envelope eventually violates the small-angle assumption; damping and nonlinear frequency shifts matter in a real pendulum.

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 mathematical model represents selected features of a system with mathematical objects and relations so that its behavior can be analyzed or predicted. Choosing the representation is part of the work: specify the quantities tracked, the equations or probability distributions connecting them, and the conditions under which those connections are expected to apply. A mathematical proof can establish a consequence of the model; comparing that consequence with observations tests its suitability for the intended use.

For a simple pendulum, the angle $\theta (t)$ and gravitational acceleration $g$ give the ideal equation $\ddot{\theta }+(g/\ell )\sin \theta =0$. Its assumptions include a point mass, fixed length $\ell$, uniform gravitational acceleration, and no damping. The small-angle approximation gives the simpler harmonic oscillator equation and oscillation period $T_0=2\pi \sqrt{\ell /g}$. A larger amplitude changes the full model's period; drag requires a damped pendulum. The simpler model remains useful when its omitted effects are small for the question being asked.

A model need not track every detail. A Bernoulli process models repeated success/failure observations using one probability and an independence assumption; a Gaussian process specifies a mean and a covariance function to model related measurements. Each representation enables calculations by retaining some distinctions and omitting others. Useful simplification therefore comes with a stated domain of use and quantities against which its predictions can be checked.

Rescaling dimensional quantities by chosen characteristic scales produces dimensionless variables. The remaining dimensionless ratios identify independent parameters and possible asymptotic regimes.

In the simple pendulum equation, choose the time scale $t_{\ast }=\sqrt{\ell /g}$ and put $\tau =t/t_{\ast }$. The chain rule gives $\ddot{\theta }=(g/\ell )d^2\theta /d{\tau }^2$, so the equation becomes $d^2\theta /d{\tau }^2+\sin \theta =0$. This removes $\ell$ and $g$ from the equation, while the initial conditions still include angular amplitude and dimensionless initial angular speed. A solution can therefore be calculated once and rescaled to many physical lengths and gravity values with the same dimensionless initial data.

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 three equal simple pendulums in a straight open chain with springs only between neighbours, the spring stiffness matrix in displacement coordinates is $k\left(\begin{array}{ccc}1& -1& 0\\ -1& 2& -1\\ 0& -1& 1\end{array}\right)$. Its eigenvectors $(1,1,1)$, $(1,0,-1)$ and $(1,-2,1)$ have respective dimensionless eigenvalues $0,1,3$. Adding the gravitational restoring term yields the displayed squared eigenfrequencies. These are free spring ends: fixed wall springs would change the matrix and mode shapes.

Each simple pendulum moves in the vertical plane containing the pivots. Ignoring contact between the ideal rods and point masses, each rod can turn through a full circle. Thus the configuration space is the torusChoose generalized coordinates ${\theta }_1,{\theta }_2$, measured from the downward vertical, positive towards the right and understood modulo $2\pi$. These angles are local coordinates on the torus, rather than a single globally one-to-one real coordinate chart. With upward Cartesian coordinate $y$, the bob positions areTheir velocities satisfy $|{\dot{\mathbf{r}}}_j{|}^2=l^2{\dot{\theta }}_j^2$, so kinetic energy from Cartesian position vectors givesThe spring length is the Euclidean distanceTaking gravitational potential energy zero at pivot height gives $V_g=-mgl(\cos {\theta }_1+\cos {\theta }_2)$. The spring potential energy is $V_s=k(d-r{)}^2/2$, since $r$ is its natural length. Consequently the Lagrangian isIn the paper's notation $t({\theta }_1,{\theta }_2)=V_s$; this function is distinct from the time variable. This is the exact Lagrangian of two spring-coupled pendulums.

One alternative Lagrangian isIt corresponds to setting gravitational potential energy zero at the downward position. The constant disappears from every Euler-Lagrange equation. More generally, total-time-derivative invariance of a Lagrangian allows $\tilde{L}=L+dF({\theta }_1,{\theta }_2,t)/dt$ for any smooth $F$.

The Euler-Lagrange equations of the quadratic Lagrangian areThe mechanical mass matrix is the identity. A normal mode $\mathbf{\theta }=\mathbf{a}\cos (\omega t+\varphi )$ therefore satisfies the eigenvalue problem $\mathsf{K}\mathbf{a}={\omega }^2\mathbf{a}$. Direct multiplication givesThese are normal modes of two spring-coupled equal pendulums. The corresponding normal mode coordinates $u={\theta }_1+{\theta }_2$ and $v={\theta }_1-{\theta }_2$ obey $\ddot{u}+u=0$ and $\ddot{v}+(1+2k)v=0$.

For the in-phase normal mode, the bobs undergo identical displacements, leaving their separation exactly $r$. The spring stays at its natural length and exerts no force, so this angular frequency is independent of $k$. The value $1$ is the individual simple pendulum angular frequency $\sqrt{g/l}$ in the chosen units.

Write $w=\sqrt{1+2k}$. The initial conditions on the normal mode coordinates are $u(0)=v(0)=0$ and $\dot{u}(0)=\dot{v}(0)=k$. Solving the two harmonic oscillator equations and recovering the original angles gives the exact solution within the harmonic approximation:Their angular velocities are ${\dot{\theta }}_1=k(\cos t+\cos wt)/2$ and ${\dot{\theta }}_2=k(\cos t-\cos wt)/2$, which verify all four initial conditions.

To expose the beat, define $\Delta =w-1$ and $\Omega =(w+1)/2$. Replacing $1/w$ by $1$ changes each angle by $O(k^2)$ uniformly in time. The sum-to-product identities then giveThe displayed angular velocities are exact in the linear system. The absolute and signed beat envelopes have leading amplitudes $k|\cos (\Delta t/2)|$ and $k|\sin (\Delta t/2)|$. Because $\Delta =k+O(k^2)$, at $\tau =\pi /k$ we have $\Delta \tau /2=\pi /2+O(k)$ and $\Omega \tau =\tau +\pi /2+O(k)$. HenceThe first simple pendulum has negligible oscillation amplitude. Its kinetic energy is $T_1(\tau )=O(k^4)$, while the second has $T_2(\tau )=k^2{\cos }^2\tau /2+O(k^3)$. Thus, to the retained order, all instantaneous kinetic energy belongs to the second simple pendulum. The fast phase can make that kinetic energy small: the whole initial energy $k^2/2$ is transferred as the second simple pendulum's combined kinetic energy and gravitational potential energy, sinceThis is energy transfer after an impulse in two weakly coupled oscillators. It does not require $\cos \tau =1$.

At $2\tau$, $\Delta (2\tau )/2=\pi +O(k)$, so the beat has returned the oscillation to the first simple pendulum:The energy localization returns, although the fast oscillation phase need not equal its initial phase. The exact first minimum of the leading beat envelope is at $\pi /\Delta$, whose leading value is $\pi /k$. Keeping the slow phase $\Delta t$ is essential: expanding $\sin (wt)$ at fixed $t$ and then using $t=O(1/k)$ would lose the energy exchange. Although angles are retained only to order $k$, their leading energies are of order $k^2$.

Assume small planar oscillations about the vertical equilibrium, with the connecting spring relaxed in that equilibrium. Let $x_j=l{\theta }_j$ be the first-order horizontal displacements of the two simple pendulums. The gravitational restoring force on bob $j$ is $-(mg/l)x_j$; the spring extension to first order is $x_2-x_1$. Thus the linearized equations of motion areAdding and subtracting them diagonalizes the system. For $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$,The normal modes and angular frequencies are thereforeThe in-phase mode does not stretch the spring, while the opposite-phase mode adds a restoring force of magnitude $2k$ times each bob's displacement. These are the small-oscillation frequencies, not exact amplitude-independent frequencies for arbitrary pendulum angles.

Write $v=\dot{x}$. The simple pendulum becomes $\dot{x}=v$, $\dot{v}=-\sin x$. Its equilibria are $(k\pi ,0)$. The linearization is $\left(\begin{array}{cc}0& 1\\ -\cos (k\pi )& 0\end{array}\right)$. For even $k$ its eigenvalues are $\pm i$, and for odd $k$ they are $\pm 1$. The conserved energyhas a strict local minimum at each even multiple of $\pi$, proving nonlinear Lyapunov stability there. The odd multiples are saddles and are unstable. ThereforeThe stable centres are not asymptotically stable because energy is conserved.

The saddle energy is $H=2$, so the separatrix is determined byEnergy levels $0<H<2$ form closed libration curves around the centres; levels $H>2$ have velocity of constant sign and describe rotations. The separatrices join neighboring saddles in the unwrapped $(x,v)$ plane. On the angle cylinder, where $x$ is identified modulo $2\pi$, they are homoclinic to the single saddle.

At $x=0$ the positive separatrix has $\dot{x}=2$. Before reaching the neighboring saddle at $x=\pi$, the velocity is $2\cos (x/2)>0$, so the time to angle $a<\pi$ isAs $a\uparrow \pi$, $\cos (x/2)\sim (\pi -x)/2$. The integrand is therefore asymptotic to $1/(\pi -x)$, whose improper integral diverges logarithmically. Thus the pendulum separatrix approach takes infinite time:Starting with velocity $-2$ gives the reflected approach to $x=-\pi$ and the same infinite time. The velocity tends to zero but never reaches zero at a finite time.

Put $I_j=m_j{r}_j^2$ and $K={\omega }^2{r}_1{r}_2$. The Euler-Lagrange equations are $I_1{\ddot{\varphi }}_1=K\sin \psi$ and $I_2{\ddot{\varphi }}_2=-K\sin \psi$, where $\psi ={\varphi }_2-{\varphi }_1$. Adding shows that $I_1{\dot{\varphi }}_1+I_2{\dot{\varphi }}_2$ is conserved. Subtracting gives the equation of a simple pendulumThe relative equilibria are $\psi =0$ and $\psi =\pi$ modulo $2\pi$. The relative energy ${\dot{\psi }}^2/2+k^2(1-\cos \psi )$ has a strict minimum at zero, proving stability there. Linearization gives $\ddot{\eta }=-k^2\eta$ near zero and $\ddot{\eta }=k^2\eta$ near $\pi$, so the latter is unstable. The small-oscillation period is $2\pi /k$. These are equilibria of the relative angle; a nonzero conserved total angular momentum permits common rotation of both particles.

Take the pivot as origin, with $x$ horizontal and $y$ vertically upward. For the simple pendulum, putThe velocity and acceleration areThe forces are the tension $-T{\mathbf{e}}_r$ and gravity $(0,-mg)$. Taking the dot product of Newton's second law with ${\mathbf{e}}_{\theta }$ eliminates the tension and givesNear the downward stable equilibrium $\theta =0$, the small-angle approximation $\sin \theta \simeq \theta$ gives $\ddot{\theta }=-(g/\ell )\theta$ to first order. Since $x=\ell \sin \theta \simeq \ell \theta$, its linearization isThis is simple harmonic motion with angular frequency $\sqrt{g/\ell }$; the restoring sign confirms the stable equilibrium.

Measure $x_1,x_2$ as horizontal displacements from the respective downward equilibrium positions; the absolute horizontal coordinates are $x_1$ and $L+x_2$. Each vertical displacement is of second order in its small angle. Thus the spring is horizontal to first order, its extension is $x_2-x_1$, and Hooke's law gives horizontal forces $k(x_2-x_1)$ and $-k(x_2-x_1)$. Combining these with the linearized simple pendulum restoring force yieldsFor the normal modes of two spring-coupled equal pendulums, add and subtract these ordinary differential equations. The normal mode coordinates $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$ satisfyHence the two natural angular frequencies and displacement patterns areOrdinary frequencies are ${\omega }_{+}/(2\pi )$ and ${\omega }_{-}/(2\pi )$. In the first normal mode, both particles swing together with equal amplitudes and the spring is unextended to first order. In the second normal mode, the particles swing in opposite directions with equal amplitudes, so the spring adds a restoring force and increases the angular frequency.

For a fixed attachment ratio $\delta >0$, making the disc much heavier than the rod givesIf the disc is also small relative to the attachment distance, $\alpha /\delta =a/d\to 0$, its moment of inertia about its own center is negligible compared with $M{d}^2$. ThusThis is a simple pendulum with a point mass at the disc's center and effective length $d$; it is length $\ell$ only if $d=\ell$. Even before the small-angle approximation, its equation tends to $\ddot{\theta }=-(g/d)\sin \theta$.

If $\delta$ is allowed to tend to zero together with the other parameters, the bare conditions $\alpha \ll 1$ and $\mu \gg 1$ do not alone justify that reduction. The needed dominance conditions are $\mu \delta \gg 1$, $\mu {\delta }^2\gg 1$, and $\alpha /\delta \ll 1$, so that the rod contributes negligible gravitational torque and moment of inertia, and the disc behaves as a point mass. They follow automatically in the combined limit with fixed $\delta >0$.