Hang two IDENTICAL pendulums and join them with a light spring; start one swinging and the other still — what does the energy do?
▶ Launch the interactive simulationIntegrate the small-angle coupled system θ¨ = −(g/L)θ − c·Δθ (velocity-Verlet), released from one pendulum at amplitude and the other at rest. Recover the slow beat envelope by low-passing pendulum 1's energy over one carrier period and timing the interval between its troughs (T_beat); read the fast carrier frequency ω̄ off pendulum 1's zero-crossings; reconstruct both normal-mode frequencies ω_s = ω̄ − Δω/2 and ω_a = ω̄ + Δω/2 and compare to theory.
the two NORMAL MODES — in-phase ω_s = √(g/L) (the spring never stretches) and out-of-phase ω_a = √(g/L + 2c) (it does) — and the beat T_beat = 2π/(ω_a − ω_s): releasing one pendulum excites an equal mix of both, so the energy drains ENTIRELY into the second pendulum and slowly returns. Because the pendulums are identical (resonant) the transfer is 100% — the left one momentarily stops dead; detune them and the swap would be incomplete. The measured beat and reconstructed mode frequencies match the closed forms to ~0.1–1%