 Energy transfer after an impulse in two weakly coupled oscillators

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.