--- title: "F-V and F-I analogy of two-mass system, X2 output" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:fv-fi-tf-m1-m2-k" source: https://www.bctengineer.com/q/semester_3/control_systems/fv-fi-tf-m1-m2-k site: BCT Engineering --- # F-V and F-I analogy of two-mass system, X2 output Control Systems · Chapter 2: Component Modeling - Asked in: 2076 Chaitra, 10 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/fv-fi-tf-m1-m2-k - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question The given mechanical system has force f(t) as input and x1 and x2 as displacement outputs. Draw equivalent f-v and f-I analogous electrical circuit and determine the transfer function with x2 as output. [Figure: fixed wall – spring k – mass M2 (displacement x2) – damper B1 – mass M1 (displacement x1, force F(t) applied); M1 is also connected to the fixed wall through damper B2] ## Short answer Write equations for M₁ and M₂, solve to get X₂(s)/F(s) = B₁ / [M₁M₂ s³ + (B₁M₁+B₁M₂+B₂M₂)s² + (B₁B₂ + kM₁)s + (B₁+B₂)k]. F–V analogy: F→e, M→L, B→R, k→1/C, i=ẋ. F–I analogy: F→i, M→C, B→1/R, k→1/L, v=ẋ; circuit has C₁=M₁, C₂=M₂, R₁=1/B₁, R₂=1/B₂, L=1/k. ## Answer Displacements: $x_1$ for $M_1$ (where $F$ acts) and $x_2$ for $M_2$. Assume no ground friction other than the dampers shown. #### Differential equations - $M_1$: inertia, damper $B_1$ (relative to $M_2$), damper $B_2$ to the wall. - $M_2$: inertia, damper $B_1$, spring $k$ to the wall. $$ \begin{aligned} M_1\ddot x_1 + B_1(\dot x_1 - \dot x_2) + B_2\dot x_1 &= F(t) \\ M_2\ddot x_2 + B_1(\dot x_2 - \dot x_1) + kx_2 &= 0 \end{aligned} $$ #### F-V analogous circuit (mesh) $F\to e$, $M\to L$, $B\to R$, $k\to 1/C$, $\dot x\to i$, $x \to q$. $$ \begin{aligned} L_1\frac{di_1}{dt} + R_1(i_1 - i_2) + R_2 i_1 &= e \\ L_2\frac{di_2}{dt} + R_1(i_2 - i_1) + \frac{1}{C}\int i_2\,dt &= 0 \end{aligned} $$ ($L_1 = M_1$, $L_2 = M_2$, $R_1 = B_1$, $R_2 = B_2$, $C = 1/k$) ```text L1=M1 R2=B2 L2=M2 +--LLLL--/\/\/--+--------LLLL--+ | | | (~) e R1 = B1 C = 1/k | i1 | i2 | +---------------+--------------+ ``` #### F-I analogous circuit (node) $F\to i$, $M\to C$, $B\to 1/R$, $k\to 1/L$, $\dot x\to v$. $$ \begin{aligned} C_1\frac{dv_1}{dt} + \frac{v_1 - v_2}{R_1} + \frac{v_1}{R_2} &= i \\ C_2\frac{dv_2}{dt} + \frac{v_2 - v_1}{R_1} + \frac{1}{L}\int v_2\,dt &= 0 \end{aligned} $$ ($C_1 = M_1$, $C_2 = M_2$, $R_1 = 1/B_1$, $R_2 = 1/B_2$, $L = 1/k$) ```text v1 R1 v2 +----+----+--/\/\/--+----+ | | | | | (i) C1 R2 C2 L | | | | | +----+----+---------+----+ gnd ``` #### Transfer function $X_2(s)/F(s)$ Laplace form: $$ \begin{aligned} \left[M_1s^2 + (B_1 + B_2)s\right]X_1 - B_1sX_2 &= F \\ -B_1sX_1 + (M_2s^2 + B_1s + k)X_2 &= 0 \end{aligned} $$ By Cramer's rule: $$ \frac{X_2}{F} = \frac{B_1s}{\left[M_1s^2 + (B_1+B_2)s\right](M_2s^2 + B_1s + k) - B_1^2s^2} $$ Cancelling the common factor $s$: $$ \frac{X_2(s)}{F(s)} = \frac{B_1}{M_1M_2s^3 + (B_1M_1 + B_1M_2 + B_2M_2)s^2 + (B_1B_2 + kM_1)s + (B_1 + B_2)k} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the equation for M2? A. M2 x2'' + B1(x2' - x1') + k x2 = 0 B. M2 x2'' + B2 x2' + k x2 = F C. M2 x2'' + k(x2 - x1) = 0 D. M2 x2'' + B1 x2' = 0 Answer: A. M2 sees damper B1 relative to M1 and the spring k to the wall; no external force. (Key point: Differential equations) 2. In the F-V mesh circuit, what represents spring k? A. Voltage source k B. Inductor L = k C. C = 1/k in mesh 2 D. Resistor k Answer: C. Springs map to capacitors C = 1/k; the spring to the wall belongs to mesh 2 only. (Key point: F-V mapping) 3. In the F-I circuit, B1 (between the masses) becomes: A. L = 1/B1 B. R1 = 1/B1 between node 1 and node 2 C. R1 = B1 to ground D. C = B1 Answer: B. Dampers become resistors 1/B, and the one joining the masses connects the two nodes. (Key point: F-I mapping) 4. Why can the factor s be cancelled in X2/F? A. The system has no poles B. Initial conditions are non-zero C. Numerator B1 s and the denominator terms all contain s D. k is zero Answer: C. Cramer's rule gives B1 s over a denominator with a common s, which cancels, leaving B1/(cubic). (Key point: Final transfer function) ## Close variants set in other exams - [Two-mass system with numeric values: F-V analogy and X1/F](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-tf-m1m2-k1k2k3) — Control Systems, chapter 2, asked 3× - [Two-mass system X1/F, X2/F with F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-tf-m1-m2-k1b1-k2b2) — Control Systems, chapter 2, asked 3× ## Related in Control Systems - [Two-mass spring system X1/F and F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-m1-m2-k1-k2) — Control Systems, chapter 2 - [Two-mass system transfer function and F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-two-mass-k1k2b1b2) — Control Systems, chapter 2 - [Two-mass system X2/F by free body diagram, F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afbd-fv-m1-m2-k1-b1) — Control Systems, chapter 2 - [Two-mass system X2/F with F-V and F-I analogy (D3)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-fi-m1-m2-2) — Control Systems, chapter 2, asked 3× - [Two-mass system X2/F with F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-m1-m2-k1k2-b1b2b3) — Control Systems, chapter 2, asked 3× - [Two-mass system X2/F with F-V analogy (B1, K3 link)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-m1-m2-b1k3-k1) — Control Systems, chapter 2