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Consider identical harmonic oscillators with unit mass, natural angular frequency ${\omega }_0>0$, and positive coupling $\kappa \ll {\omega }_0^2$:An initial impulse on the first gives $x_1(0)=x_2(0)={\dot{x}}_2(0)=0$ and ${\dot{x}}_1(0)=v$. The normal mode coordinates $u=x_1+x_2$ and $w=x_1-x_2$ have angular frequencies ${\omega }_0$ and ${\omega }_a=\sqrt{{\omega }_0^2+2\kappa }$. Solving their initial value problems provesLet $\Delta ={\omega }_a-{\omega }_0\sim \kappa /{\omega }_0$. To leading order in $\kappa /{\omega }_0^2$, sum-to-product identities give the displacement amplitude envelopes $(|v|/{\omega }_0)|\cos (\Delta t/2)|$ and $(|v|/{\omega }_0)|\sin (\Delta t/2)|$. Thus the first nearly complete energy transfer occurs at $\pi /\Delta$, and localization returns at $2\pi /\Delta$. More precisely, at $t=\pi /\Delta$, the first oscillator has exactly zero velocity and residual displacement $\frac{v}{2}(1/{\omega }_0-1/{\omega }_a)\sin ({\omega }_{0}t)$, of order $|v|\kappa /{\omega }_0^3$.

Here transferred energy means kinetic energy plus the uncoupled quadratic potential energy; the coupling potential energy is relatively small. The fast phase need not make the receiving oscillator pass through zero displacement at the transfer time. Expanding $\Delta t$ as small is invalid on the transfer scale $t=O(1/\kappa )$. This lemma complements absolute and signed beat envelopes by keeping that slow phase.

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 $k^2$.

For the normal modes of two spring-coupled equal pendulums, a normal mode has every coordinate oscillating at the same frequency with a fixed displacement ratio. The linearized equations here areTheir normalized displacement patterns and angular frequencies areThe spring is unstretched in the in-phase mode, and stretched or compressed in the opposite-phase mode. The numerical ratio follows from $k/m=(21/200){\omega }_0^2$.

The initial displacement is the equal sum of the two mode patterns, and the initial velocities vanish. ThereforeWith the printed ${\omega }_0=2\pi \mathrm{rad}{\mathrm{s}}^{-1}$ this is $x_2/a=\sin (2.1\pi t)\sin (0.1\pi t)$, with $t$ in seconds. The fast oscillation has frequency $1.05\mathrm{Hz}$ and the amplitude is bounded by $a|\sin (\pi t/10)|$. The first complete displacement transfer is at $t=5\mathrm{s}$, when $x_1=0$, $x_2=a$ and both velocities vanish; the original state returns at $10\mathrm{s}$. Distinguishing absolute and signed beat envelopes, the absolute beat envelope repeats every $10\mathrm{s}$, although its signed sine factor has period $20\mathrm{s}$.