= Normal modes of a double pendulum
{title2=$\omega_\pm^2=(g/\ell)(2\pm\sqrt2)$}
For equal <lengths> and <mass> ratio $\mu=m/M$, the squared <angular frequencies> are $\omega_0^2(1+\mu\pm\sqrt{\mu(1+\mu)})$, where $M$ is the upper <mass>, $m$ the lower <mass>, and $\omega_0^2=g/\ell$. The equal-<mass> case $\mu=1$ admits particularly simple <mode shapes>.
Use the ideal model of the <double pendulum>, with both <angles> measured from the downward vertical. Around $\theta_1=\theta_2=0$, the <Taylor expansions> $\sin\theta_i=\theta_i+O(\theta_i^3)$ and $\cos\theta_i=1-\theta_i^2/2+O(\theta_i^4)$ give the quadratic <Lagrangian>
$$
L_2=\frac12\dot{\boldsymbol\theta}^{\mathsf T}\mathsf M\dot{\boldsymbol\theta}-\frac12\boldsymbol\theta^{\mathsf T}\mathsf K\boldsymbol\theta,\qquad
\boldsymbol\theta=\begin{pmatrix}\theta_1\\\theta_2\end{pmatrix},\qquad
\mathsf M=m\ell^2\begin{pmatrix}2&1\\1&1\end{pmatrix},\quad
\mathsf K=mg\ell\begin{pmatrix}2&0\\0&1\end{pmatrix}.
$$
Here both <angles> and their <angular velocities> are counted as small. Consequently the correction to $\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2$ is fourth order and does not enter $L_2$. The <Euler-Lagrange equations> give $\mathsf M\ddot{\boldsymbol\theta}+\mathsf K\boldsymbol\theta=0$. The <mass matrix> is not diagonal: an angular motion of the upper rod carries the lower <mass> too.<Classical Dynamics by David Tong>[Classical Dynamics, chapter 2, §2.6.1, printed pp. 41–42, equations (2.130)–(2.132): equal-mass, equal-length linearized double pendulum]{ref}
For a <normal mode>, put $\boldsymbol\theta(t)=\mathbf a\cos(\omega t+\varphi)$ with fixed <amplitude> <vector> $\mathbf a$. The resulting <generalized eigenvalue problem for small oscillations> is $\mathsf K\mathbf a=\omega^2\mathsf M\mathbf a$. With $\lambda=\omega^2\ell/g$, its <characteristic equation> is
$$
0=\det\begin{pmatrix}2-2\lambda&-\lambda\\-\lambda&1-\lambda\end{pmatrix}=\lambda^2-4\lambda+2.
$$
Solving this quadratic and using the second row, $a_2/a_1=\lambda/(1-\lambda)$, gives
$$
\begin{aligned}
\omega_-^2&=\frac g\ell(2-\sqrt2),&\mathbf a_-&=\begin{pmatrix}1\\\sqrt2\end{pmatrix},\\
\omega_+^2&=\frac g\ell(2+\sqrt2),&\mathbf a_+&=\begin{pmatrix}1\\-\sqrt2\end{pmatrix}.
\end{aligned}
$$
The lower-<frequency> <normal mode> has the rods in phase, with the lower <angle> $\sqrt2$ times the upper one. The higher-<frequency> <normal mode> has opposite signs, corresponding to an <oscillation phase> difference of $\pi$. These are ratios of absolute <angles>. The horizontal <displacements> are instead $x_1\simeq\ell\theta_1$ and $x_2\simeq\ell(\theta_1+\theta_2)$; their ratios are $1+\sqrt2$ and $1-\sqrt2$ respectively. Thus the lower <mass> moves horizontally opposite to the upper <mass> in the faster <normal mode>.
\Image[/double-pendulum-linear-mode-patterns.png]
{title=Two linear normal modes: rod patterns and angle histories on the same dimensionless time scale}
{description=Original predictions for the ideal equal-<mass> <double pendulum>, with upper-<angle> <amplitude> $A=0.06$ radians and $\tau=t\sqrt{g/\ell}$. The snapshots use the rod geometry at $\boldsymbol\theta=A\mathbf a_\pm$, with horizontal offsets enlarged fourfold and heights unchanged; the curves show the <angles> predicted by <linearization>, not experimental records. Both histories use the same horizontal scale, making the faster opposed <normal mode> visible.}
Both squared <angular frequencies> are positive. Their ratio satisfies $\omega_+/\omega_-=1+\sqrt2$, independently of $m$, $g$ and $\ell$. For a constructed model with $\ell=1\,\mathrm m$ and $g=9.81\,\mathrm{m\,s^{-2}}$, the <angular frequencies> are approximately $2.3972$ and $5.7874\,\mathrm{s^{-1}}$, and the <oscillation periods> $2\pi/\omega_\pm$ are $2.6211$ and $1.0857\,\mathrm s$. Doubling both <masses> leaves these unchanged; multiplying both rod <lengths> by four doubles both <oscillation periods>.
The <eigenvectors> are not orthogonal in the ordinary <dot product>: $\mathbf a_-^{\mathsf T}\mathbf a_+=-1$. They satisfy <mass-matrix orthogonality of normal modes>, since
$$
\mathbf a_-^{\mathsf T}\mathsf M\mathbf a_+=0,\qquad
\mathbf a_-^{\mathsf T}\mathsf M\mathbf a_-=m\ell^2(4+2\sqrt2),\qquad
\mathbf a_+^{\mathsf T}\mathsf M\mathbf a_+=m\ell^2(4-2\sqrt2).
$$
Writing $\boldsymbol\theta=\mathbf a_-q_-+\mathbf a_+q_+$ therefore removes the cross terms from both quadratic <energies>. The two <normal coordinates> satisfy $\ddot q_\pm+\omega_\pm^2q_\pm=0$. This is why the <normal modes> are useful: one coupled system becomes two independently evolving <harmonic oscillators>, and any small release can be decomposed into them. At larger <angles>, the discarded terms in the <double pendulum equations of motion> matter and this exact linear <superposition> no longer applies.
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