--- title: "Two-mass system transfer function and F-V analogy" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:tf-two-mass-k1k2b1b2" source: https://www.bctengineer.com/q/semester_3/control_systems/tf-two-mass-k1k2b1b2 site: BCT Engineering --- # Two-mass system transfer function and F-V analogy Control Systems · Chapter 2: Component Modeling - Asked in: 2082 Baisakh, 8+2 marks - Video: [Translational Mechanical Systems (Solved Example)](https://www.youtube.com/watch?v=Av86LcNuiMQ) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-two-mass-k1k2b1b2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Obtain transfer function for the system as below considering displacement of mass M2 as output. Also develop Force-Voltage analogous circuit. [Figure: wall – spring K1 – mass m1 (force F applied, displacement x1, friction fc1 with ground) – spring K2 in series with damper B1 – mass m2 (displacement x2, friction fc2 with ground) – damper B2 – wall] ## Short answer X2(s)/F(s) = K2 B1 / [ (K2 + B1s)(m1 s^2 + fc1 s + K1)(m2 s + fc2 + B2) + K2 B1{(m1+m2)s^2 + (fc1+fc2+B2)s + K1} ]. In force-voltage analogy, force → voltage, mass → inductor, friction/damper → resistor, spring → capacitor. ## Answer **Assumptions:** $f_{c1}$, $f_{c2}$ are viscous friction coefficients of $m_1$, $m_2$ with the ground; $x_3$ is the displacement of the massless junction between $K_2$ and $B_1$ ($K_2$ on the $m_1$ side); zero initial conditions. #### Equations of motion (Laplace form) Mass $m_1$: $$ F(s) = (m_1s^2 + f_{c1}s + K_1)X_1 + K_2(X_1 - X_3) $$ Junction: $\;K_2(X_3 - X_1) + B_1s(X_3 - X_2) = 0$ Mass $m_2$: $$ 0 = (m_2s^2 + f_{c2}s + B_2s)X_2 + B_1s(X_2 - X_3) $$ #### Solving $K_2$ and $B_1$ in series carry the same force, so they act as one coupling element $$ Z_c(s) = \frac{K_2B_1s}{K_2 + B_1s} $$ Let $Z_1 = m_1s^2 + f_{c1}s + K_1$ and $Z_2 = m_2s^2 + (f_{c2}+B_2)s$. Then $$ \begin{bmatrix} Z_1 + Z_c & -Z_c \\ -Z_c & Z_2 + Z_c \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} F \\ 0 \end{bmatrix} $$ $$ \frac{X_2}{F} = \frac{Z_c}{(Z_1+Z_c)(Z_2+Z_c) - Z_c^2} = \frac{Z_c}{Z_1Z_2 + Z_c(Z_1 + Z_2)} $$ Substituting $Z_c$, multiplying by $(K_2 + B_1s)$ and cancelling the common factor $s$: $$ \frac{X_2(s)}{F(s)} = \frac{K_2B_1}{(K_2 + B_1s)(m_1s^2 + f_{c1}s + K_1)(m_2s + f_{c2} + B_2) + K_2B_1\left[(m_1+m_2)s^2 + (f_{c1}+f_{c2}+B_2)s + K_1\right]} $$ The denominator is a fourth-order polynomial; its leading term is $B_1m_1m_2s^4$ and its constant term is $K_1K_2(B_1 + B_2 + f_{c2})$. #### Force–voltage (F-V) analogy Force → voltage, mass → $L$, friction/damper → $R$, spring → $C = 1/K$, velocity → loop current. Mechanical series ($K_2$, $B_1$) → electrical parallel ($C_2 \parallel R_{B1}$), shared by both loops. ```text +--[L1]--[Rf1]--[C1]--+--[L2]--[Rf2]--[RB2]--+ | | | (F) i1 -> +--+--+ <- i2 | | C2 RB1 | | +--+--+ | | | | +---------------------+----------------------+ L1=m1, Rf1=fc1, C1=1/K1, C2=1/K2, RB1=B1 L2=m2, Rf2=fc2, RB2=B2 ``` Loop equations, with $i = \dot x$: $$ \begin{aligned} F &= L_1\frac{di_1}{dt} + R_{f1}i_1 + \frac{1}{C_1}\int i_1dt + v_p \\ 0 &= L_2\frac{di_2}{dt} + (R_{f2} + R_{B2})i_2 - v_p \end{aligned} $$ where $v_p$ is the voltage across $C_2 \parallel R_{B1}$ carrying $(i_1 - i_2)$ in total; these match the mechanical equations. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. K2 in series with B1 act as A. Zc(s) = K2 B1 s/(K2 + B1 s) B. K2 + B1 s C. K2 B1 D. B1 s/K2 Answer: A. Series elements share force. (Key point: Coupling element) 2. The denominator polynomial is of order A. two B. four, with leading term B1 m1 m2 s^4 C. three, with leading term m1 m2 s^3 D. five Answer: B. Two masses give s^2 each; B1 contributes one more s after cancelling. (Key point: Denominator order) 3. Mechanical series K2, B1 becomes A. series C2 and R_B1 B. a single inductor C. electrical parallel C2 || R_B1 shared by both loops D. a voltage source Answer: C. Series to parallel in F-V analogy. 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