---
title: "Two-mass system between walls with F-I analogy"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-fi-m1-m2-k1k2k3-b1b2b3"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fi-m1-m2-k1k2k3-b1b2b3
site: BCT Engineering
---

# Two-mass system between walls with F-I analogy

Control Systems · Chapter 2: Component Modeling

- Asked in: 2073 Shrawan, 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/tf-fi-m1-m2-k1k2k3-b1b2b3
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Find transfer function (consider displacement of mass M2 as output) for the given mechanical system. Also develop force-current analogous circuit. [Figure: wall – spring k1 – mass M1 (force F(t) applied, displacement x1, friction B1 with ground) – damper B2 in series with spring k2 – mass M2 (displacement x2, friction B3 with ground) – spring k3 – wall]

## Short answer

Series B2–k2 yields Keq(s) = (B2 s k2)/(B2 s + k2). With a=M1s^2+B1s+k1, b=M2s^2+B3s+k3, X2/F = Keq / [ab + Keq(a+b)]. F-I analogy: force→current, velocity→voltage, masses→C, dampers→1/R, springs→1/L.

## Answer

Displacements: $x_1$ ($M_1$), $x_2$ ($M_2$) and $x_3$ = junction between damper $B_2$ (on the $M_1$ side) and spring $k_2$ (on the $M_2$ side).

#### Differential equations
$$
\begin{aligned}
M_1\ddot x_1 + B_1\dot x_1 + k_1x_1 + B_2(\dot x_1 - \dot x_3) &= F(t) \\
B_2(\dot x_3 - \dot x_1) + k_2(x_3 - x_2) &= 0 \\
M_2\ddot x_2 + B_3\dot x_2 + k_3x_2 + k_2(x_2 - x_3) &= 0
\end{aligned}
$$

#### Eliminating the junction
The series $B_2$–$k_2$ combination acts like one element:
$$
K_{eq}(s) = \frac{B_2s\cdot k_2}{B_2s + k_2}
$$
With $a = M_1s^2 + B_1s + k_1$ and $b = M_2s^2 + B_3s + k_3$:
$$
\begin{aligned}
(a + K_{eq})X_1 - K_{eq}X_2 &= F \\
-K_{eq}X_1 + (b + K_{eq})X_2 &= 0
\end{aligned}
$$

#### Transfer function
$$
\frac{X_2}{F} = \frac{K_{eq}}{ab + K_{eq}(a + b)}
$$
Multiplying numerator and denominator by $(B_2s + k_2)$:
$$
\frac{X_2(s)}{F(s)} = \frac{B_2k_2\,s}{(B_2s + k_2)(M_1s^2 + B_1s + k_1)(M_2s^2 + B_3s + k_3) + B_2k_2s\left[(M_1+M_2)s^2 + (B_1+B_3)s + k_1 + k_3\right]}
$$
The denominator is 5th order (leading term $B_2M_1M_2s^5$, constant term $k_1k_2k_3$).

#### Force-current analogy
$F\to i$, $M\to C$, $B\to 1/R$, $k\to 1/L$, velocity $\to$ node voltage.

| Element | Analog | Connected between |
|---|---|---|
| $M_1$, $B_1$, $k_1$ | $C_1$, $1/B_1$, $1/k_1$ | node 1 and ground |
| $B_2$ | $R = 1/B_2$ | node 1 and node 3 |
| $k_2$ | $L = 1/k_2$ | node 3 and node 2 |
| $M_2$, $B_3$, $k_3$ | $C_2$, $1/B_3$, $1/k_3$ | node 2 and ground |

```text
  v1                v3                v2
 +--+--+--+--/\/\/--+--LLLL--+--+--+--+
 |  |  |  |  1/B2      1/k2  |  |  |
(i) C1 R  L                  C2 R  L
 | =M1 1/ 1/                =M2 1/ 1/
 |  |  B1 k1                 |  B3 k3
 +--+--+--+------------------+--+--+ gnd
```
Node equation at node 1: $C_1\dfrac{dv_1}{dt} + B_1v_1 + k_1\displaystyle\int v_1dt + B_2(v_1 - v_3) = i$ (conductances written directly), and similarly for nodes 3 and 2.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. x3 is between:
   A. k3 and the wall
   B. B2 and k2
   C. k1 and M1
   D. M2 and B3
   Answer: B. x3 is the massless junction of the series B2-k2 element. (Key point: Junction x3)

2. Keq of B2 and k2 in series:
   A. k2/(B2 s)
   B. B2 k2
   C. B2 s + k2
   D. B2 s k2/(B2 s + k2)
   Answer: D. Series elements: product over sum. (Key point: Equivalent element)

3. The denominator is of order:
   A. 6th
   B. 4th
   C. 5th
   D. 3rd
   Answer: C. Leading term B2 M1 M2 s^5. (Key point: Order)

4. In the F-I circuit, L = 1/k2 is connected:
   A. Across the source
   B. From node 2 to ground
   C. From node 1 to ground
   D. Between node 3 and node 2
   Answer: D. k2 joins the junction and M2, so L=1/k2 connects node 3 and node 2. (Key point: F-I connection)

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