--- title: "Two-mass system X/F with F-V and F-I analogy" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:tf-fv-fi-m1-m2-k1k2k3" source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-fi-m1-m2-k1k2k3 site: BCT Engineering --- # Two-mass system X/F with F-V and F-I analogy Control Systems · Chapter 2: Component Modeling - Asked in: 2079 Bhadra, 8+2 marks; 2076 Asoj, 10 marks; 2070 Chaitra, 6 marks (asked 3 times) - 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/tf-fv-fi-m1-m2-k1k2k3 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Find transfer function for the following mechanical system considering displacement of mass M2 as output of the system. Also develop force current analogous circuit. [Figure: wall – spring K1 – mass M1 (force f(t) applied); M1 connected to mass M2 (displacement x) through spring K2 in parallel with a series combination of damper B and spring K3] ## Short answer The transfer function is X(s)/F(s) = [B(K2+K3)s + K2K3] / Δ(s) with Δ(s) a fifth-order polynomial. The force-current analogous circuit has a current source f, capacitors M1 and M2, inductors 1/K1, 1/K2, 1/K3, and resistor 1/B, with nodes for velocities. ## Answer **Assumptions:** the masses have no friction with the ground; $x_1$ is the displacement of $M_1$, $x$ the displacement of $M_2$, and $x_3$ the displacement of the junction between damper $B$ and spring $K_3$ (B on the $M_1$ side). #### Equations of motion (Laplace, zero initial conditions) Mass $M_1$: $$ F(s) = (M_1s^2 + K_1)X_1 + K_2(X_1 - X) + Bs(X_1 - X_3) $$ Junction ($x_3$, massless): $\;Bs(X_3 - X_1) + K_3(X_3 - X) = 0$ Mass $M_2$: $\;M_2s^2X + K_2(X - X_1) + K_3(X - X_3) = 0$ #### Solving The series $B$–$K_3$ branch acts as an equivalent spring $\dfrac{BsK_3}{Bs+K_3}$. Together with $K_2$ in parallel, the coupling between the masses is $$ K_c(s) = K_2 + \frac{BK_3s}{Bs+K_3} = \frac{N(s)}{Bs+K_3}, \quad N(s) = B(K_2+K_3)s + K_2K_3 $$ The two mass equations become $(M_1s^2+K_1+K_c)X_1 - K_cX = F$ and $-K_cX_1 + (M_2s^2+K_c)X = 0$. So $$ \frac{X(s)}{F(s)} = \frac{K_c}{(M_1s^2+K_1+K_c)(M_2s^2+K_c) - K_c^2} = \frac{K_c}{(M_1s^2+K_1)(M_2s^2+K_c) + K_cM_2s^2} $$ Multiplying top and bottom by $(Bs+K_3)$ and expanding: $$ \frac{X(s)}{F(s)} = \frac{B(K_2+K_3)s + K_2K_3}{\Delta(s)} $$ $$ \begin{aligned} \Delta(s) ={}& BM_1M_2s^5 + K_3M_1M_2s^4 + B\left[K_1M_2 + (K_2+K_3)(M_1+M_2)\right]s^3 \\ &+ K_3\left[K_1M_2 + K_2(M_1+M_2)\right]s^2 + BK_1(K_2+K_3)s + K_1K_2K_3 \end{aligned} $$ #### Force–voltage (F-V) analogy Force → voltage, mass → inductance $L = M$, damper → resistance $R = B$, spring → capacitance $C = 1/K$, velocity → loop current. Mechanical parallel → electrical series and vice versa, so the coupling becomes $C_2$ in series with ($R \parallel C_3$). ```text +--[L1]--[C1]--+--[L2]--+ | | | (f) i1-> C2 <-i2 | | | | | +--+--+ | | R C3 | | +--+--+ | | | | +--------------+--------+ L1=M1, C1=1/K1, C2=1/K2, R=B, C3=1/K3, L2=M2 ``` #### Force–current (F-I) analogy Force → current source, mass → capacitance $C = M$, damper → conductance ($R = 1/B$), spring → inductance $L = 1/K$, velocity → node voltage. Each displacement point becomes a node. ```text v1 v3 v2 +---o---+---[R]------o-----[L3]---+---o | | | | | | | +----------[L2]-----------+ | | | | | (f) C1 L1 C2 | | | | +---+---+-----------------------------+ ground (reference) C1=M1, L1=1/K1, L2=1/K2, R=1/B, L3=1/K3, C2=M2 ``` Node equation at $v_1$, for example: $f = M_1\dot v_1 + K_1\!\int v_1dt + K_2\!\int(v_1-v_2)dt + B(v_1-v_3)$, which matches the force equation of $M_1$ with $v = \dot x$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The massless junction between B and K3 satisfies A. M s^2 X3 + K3 X3 = 0 B. Bs(X3 - X1) + K3(X3 - X) = 0 C. Bs X3 + K1 X3 = F D. B s X1 = K3 X Answer: B. A massless point has zero net force. (Key point: Junction equation) 2. B and K3 in series, in parallel with K2, give coupling Kc = A. K2 + B s + K3 B. K2 B K3/(Bs+K3) C. K2 + B K3 s/(Bs + K3) D. (K2+K3) B s Answer: C. Series B-K3 gives BsK3/(Bs+K3), added to K2 in parallel. (Key point: Equivalent coupling) 3. In the F-V analogy, mass, damper and spring map to A. capacitance C=M, conductance 1/B, inductance 1/K B. resistance, inductance, capacitance respectively C. L=1/M, R=1/B, C=K D. inductance L=M, resistance R=B, capacitance C=1/K Answer: D. Force-voltage: F->V, M->L, B->R, K->1/C. (Key point: F-V analogy) 4. In the F-I analogy, spring K becomes A. an inductance L = 1/K B. a capacitance C = K C. a resistance R = K D. a capacitance C = 1/K Answer: A. Force-current: M->C, B->1/R, K->1/L. (Key point: F-I analogy) ## Related in Control Systems - [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 - [Velocity transfer function of two-mass system with analogies](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-velocity-m2-fv-fi) — Control Systems, chapter 2 - [Mass inside a mass: transfer function and F-I analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fi-m1-m2-nested) — Control Systems, chapter 2 - [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 - [F-V and F-I analogy of two-mass system, X2 output](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-fi-tf-m1-m2-k) — Control Systems, chapter 2, asked 3× - [Two-mass system between walls with F-I analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fi-m1-m2-k1k2k3-b1b2b3) — Control Systems, chapter 2