---
title: "Two-mass system with numeric values: F-V analogy and X1/F"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:fv-tf-m1m2-k1k2k3"
source: https://www.bctengineer.com/q/semester_3/control_systems/fv-tf-m1m2-k1k2k3
site: BCT Engineering
---

# Two-mass system with numeric values: F-V analogy and X1/F

Control Systems · Chapter 2: Component Modeling

- Asked in: 2081 Baisakh, 2+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/fv-tf-m1m2-k1k2k3
- 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 output. Draw the equivalent F-V analogous electrical circuit and determine the transfer function with X1 as output. [Figure: wall – spring K1 = 1 N/m – mass M1 = 1 kg (displacement X1(t), friction with ground B1 = 2 Ns/m); M1 connected to mass M2 = 1 kg (displacement X2(t)) through spring K2 = 1 N/m in parallel with damper B2 = 1 Ns/m in series with spring K3 = 2 N/m; force F(t) applied to M2; M2 has friction with ground B3 = 1 Ns/m]

## Short answer

X1(s)/F(s) = (3s+2)/(s^5 + 5s^4 + 15s^3 + 20s^2 + 11s + 2). In force-voltage analogy, force → voltage, mass → inductor, damper → resistor, spring → capacitor; the coupling branch becomes a series-parallel network.

## Answer

**Assumptions:** damper $B_2$ and spring $K_3$ are in series and this pair is in parallel with $K_2$ between $M_1$ and $M_2$; $X_3$ is the junction between $B_2$ and $K_3$; zero initial conditions. Values: $M_1 = M_2 = 1$ kg, $K_1 = K_2 = 1$ N/m, $K_3 = 2$ N/m, $B_1 = 2$, $B_2 = 1$, $B_3 = 1$ N·s/m.

#### F-V analogous circuit
Force → voltage, mass → $L$, damper → $R$, spring → $C = 1/K$, velocity → loop current. The coupling ($K_2$ parallel to $B_2$–$K_3$ series) becomes $C_2$ in series with ($R_2 \parallel C_3$).

```text
  +--[L1]--[R1]--[C1]--+--[L2]--[R3]--+
  |                    |              |
  |     <- i1         C2      i2 ->  (F)
  |                    |              |
  |                 +--+--+           |
  |                 R2    C3          |
  |                 +--+--+           |
  |                    |              |
  +--------------------+--------------+
 L1=M1=1 H, R1=B1=2 Ω, C1=1/K1=1 F
 C2=1/K2=1 F, R2=B2=1 Ω, C3=1/K3=0.5 F
 L2=M2=1 H, R3=B3=1 Ω, source = F(t)
```

Loop currents $i_1 = \dot x_1$, $i_2 = \dot x_2$; the shared branch carries $(i_2 - i_1)$.

#### Equations of motion
Equivalent of series $B_2$–$K_3$: $\dfrac{B_2sK_3}{B_2s+K_3} = \dfrac{2s}{s+2}$. Total coupling:

$$
K_c(s) = 1 + \frac{2s}{s+2} = \frac{3s+2}{s+2}
$$

$$
\begin{aligned}
M_1: &\quad (s^2 + 2s + 1)X_1 + K_c(X_1 - X_2) = 0 \\
M_2: &\quad (s^2 + s)X_2 + K_c(X_2 - X_1) = F(s)
\end{aligned}
$$

#### Transfer function $X_1/F$
From the first equation, $X_2 = \dfrac{s^2+2s+1+K_c}{K_c}X_1$. Substituting in the second:

$$
\frac{X_1}{F} = \frac{K_c}{(s^2+2s+1)(s^2+s) + K_c(2s^2+3s+1)}
$$

Using $(s^2+2s+1)(s^2+s) = s(s+1)^3$ and $2s^2+3s+1 = (2s+1)(s+1)$, then multiplying top and bottom by $(s+2)$:

$$
\begin{aligned}
\frac{X_1}{F} &= \frac{3s+2}{s(s+2)(s+1)^3 + (3s+2)(2s+1)(s+1)} \\
&= \frac{3s+2}{(s+1)\left[s(s+2)(s+1)^2 + (3s+2)(2s+1)\right]} \\
&= \frac{3s+2}{(s+1)(s^4 + 4s^3 + 11s^2 + 9s + 2)}
\end{aligned}
$$

**Answer:**

$$
\frac{X_1(s)}{F(s)} = \frac{3s+2}{s^5 + 5s^4 + 15s^3 + 20s^2 + 11s + 2}
$$

## Check yourself

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

1. In the F-V analogy used here, which mapping is correct?
   A. Mass -> C, spring -> L, damper -> R
   B. Mass -> L, spring -> C = 1/K, damper -> R, velocity -> loop current
   C. Mass -> L, spring -> C = K, damper -> 1/R
   D. Mass -> R, spring -> L, damper -> C
   Answer: B. Force-voltage: force is voltage, mass is inductance, damper is resistance, spring is capacitance 1/K and velocity is loop current. (Key point: F-V element mapping)

2. B2 and K3 are in series between the masses. Their equivalent stiffness in the s-domain is:
   A. B2 s K3
   B. (B2 s + K3) / (B2 s K3)
   C. B2 s K3 / (B2 s + K3), which is 2s/(s+2) here
   D. B2 s + K3
   Answer: C. Elements in series carry the same force, so their impedances combine like a product over sum: B2 s K3/(B2 s + K3) = 2s/(s+2). (Key point: Series damper-spring equivalent)

3. K2 is in parallel with the B2-K3 series pair. What is the total coupling Kc(s)?
   A. 2s/(s+2)
   B. 1 + 2s + 2
   C. 1/(1 + 2s/(s+2))
   D. (3s+2)/(s+2)
   Answer: D. Parallel elements share the same displacement difference, so stiffnesses add: 1 + 2s/(s+2) = (3s+2)/(s+2). (Key point: Total coupling Kc(s))

4. What is the final transfer function X1(s)/F(s)?
   A. (3s+2)/(s^5 + 4s^4 + 11s^3 + 9s^2 + 2s)
   B. (3s+2)/(s^4 + 4s^3 + 11s^2 + 9s + 2)
   C. (s+2)/(s^5 + 5s^4 + 15s^3 + 20s^2 + 11s + 2)
   D. (3s+2)/(s^5 + 5s^4 + 15s^3 + 20s^2 + 11s + 2)
   Answer: D. Multiplying (s+1) by (s^4+4s^3+11s^2+9s+2) gives the fifth-order denominator, with numerator 3s+2. (Key point: Final X1/F result)

## Close variants set in other exams

- [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 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×

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