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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.

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 equal rod lengths $l$, upper mass $m$ and lower mass $\alpha m>0$, use horizontal displacements $x_1=l{\theta }_1$, $x_2=l({\theta }_1+{\theta }_2)$ at first order. The quadratic kinetic energy is $m{\dot{x}}_1^2/2+\alpha m{\dot{x}}_2^2/2$, while the stiffness matrix isThe generalized eigenvalue problem for small oscillations is $Ka={\omega }^{2}Ma$, not $Ka=m{\omega }^{2}a$. In particular the lower Euler-Lagrange equation is ${\ddot{x}}_2=-(g/l)(x_2-x_1)$, independent of the lower mass factor after division by that mass. This reproduces the normal modes of a double pendulum and distinguishes the mechanical mass matrix from its stiffness matrix.

A normal mode, also called an eigenmode, is a pattern of free oscillation in a linear system whose relative displacements stay fixed while a single temporal factor changes their size and sign. For an undamped stable vibrational mode,Here $\varphi$ is the mode shape, $A$ sets the amplitude, $\omega$ is the eigenfrequency expressed as an angular frequency, and $\phi$ is the oscillation phase. Every snapshot is a scalar multiple of the same mode shape; neighbouring regions may move in opposite directions. Normal modes of a double pendulum show that normal modes also describe coupled discrete systems, beyond strings. Their shapes are eigenvectors, rather than spatial eigenfunctions. ^[ref]

For the ideal wave equation on a string, take a uniform string of length $L$ with constant tension $\tau$ and linear mass density $\mu$. Assume small transverse slope, negligible bending stiffness, no damping, and fixed endpoints. Its wave speed is $c=\sqrt{\tau /\mu }$, and the nonzero normal modes areThe angular frequency ${\omega }_n$ is in radians per second; the frequency $f_n$ is in hertz. The wavelength is $2L/n$. The mode shape has $n$ loops separated by nodes of a standing wave, including both fixed endpoints. ^[ref]

As a worked example, choose $L=1\mathrm{m}$, $\tau =1\mathrm{N}$, and $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$. Then $c=1\mathrm{m}{\mathrm{s}}^{-1}$ and $f_1=0.5\mathrm{Hz}$, $f_2=1\mathrm{Hz}$, $f_3=1.5\mathrm{Hz}$. Their oscillation periods are $T_1=2\mathrm{s}$, $T_2=1\mathrm{s}$, and $T_3=2/3\mathrm{s}$. These are deliberately simple model parameters, rather than measurements of a musical instrument. The plot divides displacement by $A$ so that the three mode shapes can be compared, and uses each mode's own oscillation period for its snapshots.

For $\phi =0$, at $t=T_n/4$ every point has zero displacement, but most points have nonzero velocity. Only zeros of the mode shape are permanent nodes of a standing wave. At $t=T_n/2$ the original mode shape has reversed sign, and at $t=T_n$ it returns. The synchronized conversion between kinetic energy and potential energy is shown in normal-mode energy exchange.

The short derivation uses separation of variables: substituting $u=\varphi q$ into the wave equation gives ${\varphi }^{\prime \prime }+k^2\varphi =0$ and $q^{\prime \prime }+c^2{k}^{2}q=0$. The fixed-end boundary conditions remove the cosine spatial term and require $\sin (kL)=0$, hence $k=n\pi /L$ and ${\omega }_n=ck$. Different geometries and boundary conditions give different eigenfunctions; a sine mode shape is a property of this uniform fixed string, not a universal form of an eigenmode.

The usefulness is that a complicated vibration becomes a collection of harmonic oscillators. For sufficiently regular fixed-end initial data, the Fourier sine series givesThe initial displacement and velocity determine $a_n$ and $b_n$. This is the superposition principle in a Fourier sine basis, and modal energy distribution measures how strongly each normal mode is excited. The striking simplification is that each component keeps its identity while their sum can continually change shape. ^[ref]

These independent normal modes belong to the linear model. At larger slopes, changing tension introduces nonlinear effects and can couple the modes. A linearly damped string still admits a Fourier sine basis, but its temporal factors decay and need not oscillate; its energy dissipation identity for a linearly damped string replaces conservation of energy. More general damping need not leave undamped mode shapes independent.

First solve the supplied equations as a linear system, with ${\omega }_0^2=g/l$ and $\alpha >0$. A normal mode $\mathbf{x}=\mathbf{a}{e}^{i\omega t}$ obeysIts characteristic polynomial is ${\lambda }^2-(1+3\alpha )\lambda +\alpha (1+\alpha )$. HenceBoth are positive, since the trace and determinant of $D$ are positive.

There is a physical discrepancy in the source. The horizontal-coordinate mass matrix of an unequal-mass double pendulum follows from the stated horizontal displacements, which giveThe Euler-Lagrange equations reproduce the supplied upper equation, but give ${\ddot{x}}_2=-(g/l)(x_2-x_1)$, without its extra $\alpha$. The true normal modes of a double pendulum consequently have ${\omega }_{\pm }^2=(g/l)[1+\alpha \pm \sqrt{\alpha (1+\alpha )}]$. The boxed answer and the remaining parts below concern the explicitly supplied linear system. The two models agree at $\alpha =1$, but not for a general mass ratio.