= Energy transfer after an impulse in two weakly coupled oscillators
{title2=$t_{\mathrm{transfer}}\sim\pi\omega_0/\kappa$}
Consider identical <harmonic oscillators> with unit <mass>, natural <angular frequency> $\omega_0>0$, and positive coupling $\kappa\ll\omega_0^2$:
$$
\ddot x_1+\omega_0^2x_1+\kappa(x_1-x_2)=0,\qquad
\ddot x_2+\omega_0^2x_2+\kappa(x_2-x_1)=0.
$$
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> proves
$$
x_1=\frac v2\left(\frac{\sin\omega_0t}{\omega_0}+\frac{\sin\omega_at}{\omega_a}\right),\qquad
x_2=\frac v2\left(\frac{\sin\omega_0t}{\omega_0}-\frac{\sin\omega_at}{\omega_a}\right).
$$
Let $\Delta=\omega_a-\omega_0\sim\kappa/\omega_0$. 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> $\tfrac v2(1/\omega_0-1/\omega_a)\sin(\omega_0t)$, 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.
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