--- title: "Two-mass system equations, X2/F, F-I and F-V analogies" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:de-tf-fi-fv-m1-m2" source: https://www.bctengineer.com/q/semester_3/control_systems/de-tf-fi-fv-m1-m2 site: BCT Engineering --- # Two-mass system equations, X2/F, F-I and F-V analogies Control Systems · Chapter 2: Component Modeling - Asked in: 2069 Chaitra, 8+4 marks - Video: [2.Force Current and Force Voltage Analogy solved problem with system equation](https://www.youtube.com/watch?v=jgCfYudsIFY) — ELECTRICAL HUB - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/de-tf-fi-fv-m1-m2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Write differential equations governing the mechanical system shown in figure below and find X2(s)/F(s). Also tabulating the necessary analogies draw the Force-Current and Force-Voltage electrical analogous circuit. [Figure: force f(t) applied to mass M1 (displacement x1, friction B1 with ground); M1 connected to mass M2 (displacement x2, on rollers) through damper B12 in parallel with spring K12; M2 connected to the fixed wall through damper B2 in parallel with spring K2] ## Short answer M1 x1'' + B1 x1' + B12(x1'-x2') + K12(x1-x2)=f, M2 x2'' + B12(x2'-x1') + K12(x2-x1) + B2 x2' + K2 x2=0. X2/F = (B12 s + K12)/Δ, Δ = [M1 s^2+(B1+B12)s+K12][M2 s^2+(B12+B2)s+K12+K2] - (B12 s+K12)^2. F-V: force→voltage, mass→L, damper→R, spring→1/C; F-I: force→current, mass→C, damper→1/R, spring→1/L. ## Answer Displacements: $x_1$ for $M_1$ (force $f$, ground friction $B_1$) and $x_2$ for $M_2$ (on rollers, so no ground friction). $B_{12} \parallel K_{12}$ couples the masses; $B_2 \parallel K_2$ ties $M_2$ to the wall. #### Differential equations $$ \begin{aligned} M_1\ddot x_1 + B_1\dot x_1 + B_{12}(\dot x_1 - \dot x_2) + K_{12}(x_1 - x_2) &= f(t) \\ M_2\ddot x_2 + B_{12}(\dot x_2 - \dot x_1) + K_{12}(x_2 - x_1) + B_2\dot x_2 + K_2x_2 &= 0 \end{aligned} $$ #### Transfer function $X_2(s)/F(s)$ Laplace form: $$ \begin{aligned} \left[M_1s^2 + (B_1 + B_{12})s + K_{12}\right]X_1 - (B_{12}s + K_{12})X_2 &= F \\ -(B_{12}s + K_{12})X_1 + \left[M_2s^2 + (B_{12} + B_2)s + K_{12} + K_2\right]X_2 &= 0 \end{aligned} $$ From the second equation, $X_1 = \dfrac{M_2s^2 + (B_{12}+B_2)s + K_{12} + K_2}{B_{12}s + K_{12}}X_2$. Substituting into the first: $$ \frac{X_2(s)}{F(s)} = \frac{B_{12}s + K_{12}}{\Delta} $$ $$ \Delta = \left[M_1s^2 + (B_1+B_{12})s + K_{12}\right]\left[M_2s^2 + (B_{12}+B_2)s + K_{12} + K_2\right] - (B_{12}s + K_{12})^2 $$ Expanded: $$ \begin{aligned} \Delta =\; & M_1M_2s^4 + \left[B_1M_2 + B_{12}(M_1 + M_2) + B_2M_1\right]s^3 \\ & + \left[B_1B_{12} + B_1B_2 + B_{12}B_2 + K_{12}(M_1 + M_2) + K_2M_1\right]s^2 \\ & + \left[B_1(K_{12} + K_2) + B_{12}K_2 + B_2K_{12}\right]s + K_{12}K_2 \end{aligned} $$ #### Table of analogies | Mechanical (translational) | F-V (mesh) | F-I (node) | |---|---|---| | Force $f$ | Voltage $e$ | Current $i$ | | Mass $M$ | Inductance $L$ | Capacitance $C$ | | Damper $B$ | Resistance $R$ | Conductance $1/R$ | | Spring $K$ | Elastance $1/C$ | Reciprocal inductance $1/L$ | | Displacement $x$ | Charge $q$ | Flux linkage $\psi$ | | Velocity $\dot x$ | Current $i$ | Voltage $v$ | | Elements in parallel (same $\dot x$) | Series | Parallel | #### Force-current analogous circuit Node 1 ↔ $\dot x_1$, node 2 ↔ $\dot x_2$; $C_1 = M_1$, $C_2 = M_2$, $R_1 = 1/B_1$, $R_{12} = 1/B_{12}$, $L_{12} = 1/K_{12}$, $R_2 = 1/B_2$, $L_2 = 1/K_2$. ```text v1 L12∥R12 v2 +----+----+---+--LLLL--+---+----+----+ | | | +-/\/\/--+ | | | (i) C1 R1 C2 R2 L2 | | | | | | +----+----+----------------+----+----+ gnd ``` $$ \begin{aligned} C_1\frac{dv_1}{dt} + \frac{v_1}{R_1} + \frac{v_1 - v_2}{R_{12}} + \frac{1}{L_{12}}\int(v_1 - v_2)dt &= i \\ C_2\frac{dv_2}{dt} + \frac{v_2 - v_1}{R_{12}} + \frac{1}{L_{12}}\int(v_2 - v_1)dt + \frac{v_2}{R_2} + \frac{1}{L_2}\int v_2dt &= 0 \end{aligned} $$ #### Force-voltage analogous circuit Mesh 1 ↔ $\dot x_1$, mesh 2 ↔ $\dot x_2$; $L_1 = M_1$, $L_2 = M_2$, $R_1 = B_1$, $R_{12} = B_{12}$, $C_{12} = 1/K_{12}$, $R_2 = B_2$, $C_2 = 1/K_2$. ```text L1 R1 L2 R2 C2 +-LLLL-/\/\-+----------+-LLLL-/\/\---||--+ | | | (~) e R12 | | | | | i1 C12 i2 | +-----------+----------------------------+ ``` $$ \begin{aligned} L_1\frac{di_1}{dt} + R_1i_1 + R_{12}(i_1 - i_2) + \frac{1}{C_{12}}\int(i_1 - i_2)dt &= e \\ L_2\frac{di_2}{dt} + R_{12}(i_2 - i_1) + \frac{1}{C_{12}}\int(i_2 - i_1)dt + R_2i_2 + \frac{1}{C_2}\int i_2dt &= 0 \end{aligned} $$ Both sets have the same form as the mechanical equations, confirming the analogies. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the M2 equation have no ground-friction term? A. M2 is on rollers B. M2 has no mass C. M2 is attached to the wall D. The damper B12 replaces it Answer: A. M2 rests on rollers so no friction acts on it, and only B12, K12, B2 and K2 appear. (Key point: Equation for M2) 2. What is the numerator of X2(s)/F(s)? A. B2 s + K2 B. B12 s + K12 C. M2 s^2 + B2 s D. K12 K2 Answer: B. F acts on M1 and reaches M2 through the coupling B12 || K12, which gives B12 s + K12. (Key point: Numerator of X2/F) 3. In the F-I analogy, what do mass, damper and spring become? A. L, R, C B. C, conductance 1/R, 1/L C. R, C, L D. C, R, L Answer: B. F-I: M to C, B to 1/R, K to 1/L, force to current, velocity to voltage. (Key point: F-I analogy) 4. In the F-V circuit, how does B12 in parallel with K12 appear? A. As a parallel R and C pair in mesh 1 only B. As a series R12 and C12 in the branch shared by both meshes C. As two separate meshes D. As an inductor Answer: B. Mechanical parallel elements share the same velocity difference, so they become series electrical elements in the common branch. 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