Electronic Equivalent of a Mechanical Impact Oscillator

The paper derives the circuit equations (3)–(4), eliminates V_q to obtain the closed second‑order ODE for V_s(t) (7), maps x = −V_s and identifies R_3C_2 = R_8C_1 = 1/ω, R_2/(2R_6) = ζ, R_7 = R_2 = R_1, and V_3(t) = V_A cos(Ωt) to recover the mechanical model (8)–(9), and implements impacts by flipping the sign of V_q, which yields R = 1 for the restitution law (2b) . The candidate solution performs the same steps (differentiate (4b), use (4a) and (3), define x := −V_s, non-dimensionalize, and show that V_q sign flip implies ẋ jump), adding an explicit demonstration that V_q(t_c^+) = −V_q(t_c^−) gives ẋ(τ_c^+) = −ẋ(τ_c^−), i.e., R = 1. The only issues are minor: a sign inconsistency between (3) and (5b) in the paper’s transcription, and a typographical “R_3C_2 = R_8C_1 = ω^{-1} = 1” that should read R_3C_2 = R_8C_1 = ω^{-1}. These do not affect the main equivalence result. Overall, both are correct and essentially the same proof.