--- title: "Two-mass system X2/F and F-I analogy" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:fi-tf-two-mass-k1k2b1b2" source: https://www.bctengineer.com/q/semester_3/control_systems/fi-tf-two-mass-k1k2b1b2 site: BCT Engineering --- # Two-mass system X2/F and F-I analogy Control Systems · Chapter 2: Component Modeling - Asked in: 2080 Baisakh, 8 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/fi-tf-two-mass-k1k2b1b2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Find the transfer X2(s)/F(s) for the given mechanical system. Also develop F-I analogy circuit. [Figure: wall – spring K1 – mass M (force f applied, displacement X1, friction fc1 with ground) – spring K2 in series with damper B1 – mass M (displacement X2, friction fc2 with ground) – damper B2 – wall] ## Short answer Introduce a massless junction x₃ between K₂ and B₁, write equations, and eliminate x₃ with equivalent stiffness K_eq = K₂B₁s/(K₂+B₁s). Solving yields X₂(s)/F(s) = B₁K₂ / [(K₂+B₁s)(M₁s²+f_c₁s+K₁)(M₂s+f_c₂+B₂) + B₁K₂((M₁+M₂)s²+(f_c₁+f_c₂+B₂)s+K₁)]. F–I analogy: f→i, M→C, B→1/R, K→1/L; nodes v₁, v₂; C₁=M₁, C₂=M₂ to ground, L₁=1/K₁, L₂=1/K₂, R_B1=1/B₁. ## Answer Let $x_1$, $x_2$ be the displacements of the two masses ($M_1$, $M_2$) and $x_3$ the displacement of the massless junction between $K_2$ and $B_1$ (spring on the $M_1$ side). #### Differential equations $$ \begin{aligned} M_1\ddot x_1 + f_{c1}\dot x_1 + K_1 x_1 + K_2(x_1 - x_3) &= f \\ K_2(x_3 - x_1) + B_1(\dot x_3 - \dot x_2) &= 0 \\ M_2\ddot x_2 + (f_{c2} + B_2)\dot x_2 + B_1(\dot x_2 - \dot x_3) &= 0 \end{aligned} $$ #### Eliminating the junction From the second equation, $X_3 = \dfrac{K_2X_1 + B_1sX_2}{K_2 + B_1 s}$. The force through the series spring–damper is then $$ K_2(X_1 - X_3) = K_{eq}(X_1 - X_2), \qquad K_{eq}(s) = \frac{K_2 B_1 s}{K_2 + B_1 s} $$ So, writing $a = M_1s^2 + f_{c1}s + K_1$ and $b = M_2s^2 + (f_{c2}+B_2)s$: $$ \begin{aligned} (a + K_{eq})X_1 - K_{eq}X_2 &= F \\ -K_{eq}X_1 + (b + K_{eq})X_2 &= 0 \end{aligned} $$ #### Transfer function $$ \frac{X_2(s)}{F(s)} = \frac{K_{eq}}{(a+K_{eq})(b+K_{eq}) - K_{eq}^2} = \frac{K_{eq}}{ab + K_{eq}(a+b)} $$ Multiplying through by $(K_2 + B_1 s)$ and cancelling the common factor $s$: $$ \frac{X_2(s)}{F(s)} = \frac{B_1K_2}{(K_2 + B_1s)(M_1s^2 + f_{c1}s + K_1)(M_2s + f_{c2} + B_2) + B_1K_2\left[(M_1+M_2)s^2 + (f_{c1}+f_{c2}+B_2)s + K_1\right]} $$ The denominator is a 4th-order polynomial in $s$. #### F-I analogy $f \to i$, $M \to C$, friction/damper $B \to$ conductance $1/R$, $K \to 1/L$, velocity $\to$ node voltage. Nodes $v_1, v_3, v_2$ correspond to $\dot x_1, \dot x_3, \dot x_2$. | Mechanical | Electrical (F-I) | |---|---| | $M_1$, $M_2$ | $C_1 = M_1$, $C_2 = M_2$ (node to ground) | | $f_{c1}$, $f_{c2}$, $B_2$ | $R = 1/f_{c1}$, $1/f_{c2}$, $1/B_2$ (node to ground) | | $K_1$ | $L_1 = 1/K_1$ (node 1 to ground) | | $K_2$ | $L_2 = 1/K_2$ (node 1 to node 3) | | $B_1$ | $R_{B1} = 1/B_1$ (node 3 to node 2) | ```text v1 L2 v3 RB1 v2 +----+----+----+--LLLL--+--/\/\/--+----+----+ | | | | | | | (i) C1 R1 L1 C2 R2 R3 | | | | | | | +----+----+----+------------------+----+----+ gnd ``` ($C_1 = M_1$, $R_1 = 1/f_{c1}$, $L_1 = 1/K_1$; $L_2 = 1/K_2$, $R_{B1} = 1/B_1$; $C_2 = M_2$, $R_2 = 1/f_{c2}$, $R_3 = 1/B_2$.) Node equation at node 1: $C_1\dfrac{dv_1}{dt} + \dfrac{v_1}{R_1} + \dfrac{1}{L_1}\int v_1dt + \dfrac{1}{L_2}\int (v_1 - v_3)dt = i$, and similarly for nodes 3 and 2 (matching the three mechanical equations). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the role of x3 and how is it removed? A. It is a third mass; keep it in the final answer B. It is the input; differentiate it away C. It is the wall; set it to zero D. It is a massless junction between K2 and B1; solve its force-balance equation for X3 and substitute Answer: D. X3 has no mass, so its equation K2(X3-X1)+B1 s(X3-X2)=0 gives X3 in terms of X1 and X2. (Key point: Eliminating the junction) 2. The series spring K2 and damper B1 behave as one element Keq(s) equal to: A. K2 B1 s / (K2 + B1 s) B. (K2 + B1 s) / (K2 B1 s) C. K2 + B1 s D. K2 / (B1 s) Answer: A. Series elements carry the same force, so Keq = K2 B1 s/(K2 + B1 s). (Key point: Equivalent series element) 3. What order is the denominator of X2/F? A. 6th B. 2nd C. 4th D. 3rd Answer: C. After clearing (K2 + B1 s), two second-order masses plus the first-order junction term give a 4th-order polynomial. (Key point: Order of the denominator) 4. In the F-I circuit, how is K2 (between M1 and the junction) represented? A. Resistor 1/K2 from node 3 to node 2 B. Inductor L2 = 1/K2 between node 1 and node 3 C. Capacitor K2 from node 1 to ground D. Inductor L = K2 to ground Answer: B. Springs map to inductors 1/K; a spring joining two masses/junctions is connected between their nodes. 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