--- title: "Velocity transfer function of two-mass system with analogies" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:tf-velocity-m2-fv-fi" source: https://www.bctengineer.com/q/semester_3/control_systems/tf-velocity-m2-fv-fi site: BCT Engineering --- # Velocity transfer function of two-mass system with analogies Control Systems · Chapter 2: Component Modeling - Asked in: 2080 Baisakh, 8 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-velocity-m2-fv-fi - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Find the transfer function of mechanical system given below taking output as velocity of mass M2; also draw F-V and F-I analogy electrical networks. [Figure: left wall – damper B1 – mass m1 (displacement x1) – spring k1 – mass m2 (displacement x2, force f(t) applied); m2 also connected to the left wall through spring k2; m2 connected to the right wall through damper B2 and spring k3 in parallel] ## Short answer Solve equations for masses, X2/F = (m1s^2+B1s+k1)/Δ with 4th-order Δ. Velocity output: V2(s)/F(s) = s(m1s^2+B1s+k1)/Δ. F-V analogy: e for F, L for m, R for B, C=1/k; k2 & k3 act as series capacitors 1/(k2+k3). F-I analogy uses current source, C for m, 1/L for k; output V2 is node 2 voltage. ## Answer Displacements $x_1$ ($m_1$) and $x_2$ ($m_2$, force $f$). Output: velocity of $m_2$, $v_2 = \dot x_2$, so $V_2(s) = sX_2(s)$. #### Equations of motion - $m_1$: damper $B_1$ to the left wall, spring $k_1$ to $m_2$. - $m_2$: spring $k_1$ to $m_1$, $k_2$ to the left wall, $B_2 \parallel k_3$ to the right wall. $$ \begin{aligned} m_1\ddot x_1 + B_1\dot x_1 + k_1(x_1 - x_2) &= 0 \\ m_2\ddot x_2 + B_2\dot x_2 + (k_2 + k_3)x_2 + k_1(x_2 - x_1) &= f(t) \end{aligned} $$ Laplace form: $$ \begin{aligned} (m_1s^2 + B_1s + k_1)X_1 - k_1X_2 &= 0 \\ -k_1X_1 + (m_2s^2 + B_2s + k_1 + k_2 + k_3)X_2 &= F \end{aligned} $$ #### Transfer function $$ \frac{X_2}{F} = \frac{m_1s^2 + B_1s + k_1}{\Delta}, \qquad \Delta = (m_1s^2 + B_1s + k_1)(m_2s^2 + B_2s + k_1 + k_2 + k_3) - k_1^2 $$ Velocity output: $$ \frac{V_2(s)}{F(s)} = \frac{s(m_1s^2 + B_1s + k_1)}{\Delta} $$ with $$ \begin{aligned} \Delta =\; & m_1m_2s^4 + (B_1m_2 + B_2m_1)s^3 + \left[B_1B_2 + m_1(k_1 + k_2 + k_3) + m_2k_1\right]s^2 \\ & + \left[B_1(k_1 + k_2 + k_3) + B_2k_1\right]s + k_1(k_2 + k_3) \end{aligned} $$ #### F-V analogy (mesh, velocity $\to$ current) ```text L1=m1 R1=B1 L2=m2 R2=B2 C=1/(k2+k3) +-LLLL---/\/\--+----------LLLL----/\/\------||-----+ | | | | C1=1/k1 (~) e | i1 | i2 | +--------------+-----------------------------------+ ``` ($k_2$ and $k_3$ both act on $x_2$ alone, so they combine into one capacitor $1/(k_2+k_3)$, i.e. $C_2 = 1/k_2$ and $C_3 = 1/k_3$ in series.) The required output $V_2$ corresponds to mesh current $i_2$. #### F-I analogy (node, velocity $\to$ voltage) ```text v1 L1 v2 +----+--LLLL--+----+----+----+----+ | | | | | | | C1 R1 C2 R2 L2 L3 (i) | | | | | | | +----+--------+----+----+----+----+ gnd ``` ($C_1 = m_1$, $R_1 = 1/B_1$, $L_1 = 1/k_1$, $C_2 = m_2$, $R_2 = 1/B_2$, $L_2 = 1/k_2$, $L_3 = 1/k_3$.) The output velocity $v_2$ is the voltage of node 2. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With output as the velocity of m2, how is V2(s)/F(s) obtained from X2(s)/F(s)? A. Multiply X2/F by s B. Divide X2/F by s C. Multiply X2/F by s^2 D. Leave it unchanged Answer: A. v2 = dx2/dt, so V2(s) = sX2(s); the transfer function is X2/F multiplied by s. (Key point: Velocity output) 2. In the equation of motion of m2, which spring terms appear with x2 alone (not x2 - x1)? A. k2 and k3 B. k1 only C. k1 and k2 D. k1 and k3 Answer: A. k2 (to the left wall) and k3 (to the right wall) act on x2 only; k1 connects m1 and m2 so it gives k1(x2 - x1). (Key point: Spring terms on m2) 3. The numerator of V2(s)/F(s) is s(m1 s^2 + B1 s + k1). Where does the factor (m1 s^2 + B1 s + k1) come from? A. Eliminating X1 using the m1 equation B. The m2 equation only C. The damper B2 D. The spring k3 Answer: A. The m1 equation gives X1 = k1X2/(m1s^2+B1s+k1); substituting into the m2 equation leaves this factor as numerator. (Key point: Numerator of the result) 4. In the F-V (force-voltage) analogy, how are k2 and k3 represented? A. Capacitors 1/k2 and 1/k3 in series, giving one capacitor 1/(k2+k3) B. Inductors in parallel C. Resistors in series D. Capacitors 1/k2 and 1/k3 in parallel Answer: A. Springs map to capacitors C = 1/k; springs sharing the same displacement add stiffness, so k2+k3 equals series capacitors with C = 1/(k2+k3). 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