---
title: "Two-mass system X2/F with F-V analogy (B1, K3 link)"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-fv-m1-m2-b1k3-k1"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-m1-m2-b1k3-k1
site: BCT Engineering
---

# Two-mass system X2/F with F-V analogy (B1, K3 link)

Control Systems · Chapter 2: Component Modeling

- Asked in: 2080 Bhadra, 8+2 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-fv-m1-m2-b1k3-k1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Find the transfer function X2(s)/F(s). Also draw the F-V analogous circuit. [Figure: force F applied to mass M1 (displacement x1, ground friction Bv1); M1's right side connected through damper B1 in series with spring K3 to mass M2 (displacement x2, ground friction Bv2); M1 also connected through spring K1 to the ground/fixed frame; M2 connected to the right wall through spring K2 and through damper B2 to the ground/fixed frame]

## Short answer

The series B1−K3 reduces to Keq = B1K3s/(B1s+K3). Then X2(s)/F(s) = Keq/[ab+Keq(a+b)] where a = M1s^2+Bv1s+K1, b = M2s^2+(Bv2+B2)s+K2; denominator is 5th order with zero at s=0. F-V analogy: F→e, B1-K3 branch becomes parallel R(B1)||C(1/K3) in the shared mesh.

## Answer

Displacements: $x_1$ ($M_1$), $x_2$ ($M_2$) and $x_3$ = junction between damper $B_1$ (on the $M_1$ side) and spring $K_3$ (on the $M_2$ side).

#### Equations of motion (Laplace)
$$
\begin{aligned}
(M_1s^2 + B_{v1}s + K_1)X_1 + B_1s(X_1 - X_3) &= F \\
B_1s(X_3 - X_1) + K_3(X_3 - X_2) &= 0 \\
\left[M_2s^2 + (B_{v2} + B_2)s + K_2\right]X_2 + K_3(X_2 - X_3) &= 0
\end{aligned}
$$

#### Eliminating the junction
$B_1$ in series with $K_3$:
$$
K_{eq}(s) = \frac{B_1sK_3}{B_1s + K_3}
$$
With $a = M_1s^2 + B_{v1}s + K_1$ and $b = M_2s^2 + (B_{v2} + B_2)s + K_2$:
$$
\frac{X_2}{F} = \frac{K_{eq}}{(a + K_{eq})(b + K_{eq}) - K_{eq}^2} = \frac{K_{eq}}{ab + K_{eq}(a + b)}
$$

#### Transfer function
Multiplying by $(B_1s + K_3)$:
$$
\frac{X_2(s)}{F(s)} = \frac{B_1K_3s}{(B_1s + K_3)(M_1s^2 + B_{v1}s + K_1)\left[M_2s^2 + (B_{v2} + B_2)s + K_2\right] + B_1K_3s\left[(M_1 + M_2)s^2 + (B_{v1} + B_{v2} + B_2)s + K_1 + K_2\right]}
$$
The denominator is 5th order (leading term $B_1M_1M_2s^5$, constant term $K_1K_2K_3$). The zero at $s = 0$ shows that a steady force cannot move $M_2$, because the damper $B_1$ transmits no static force.

#### F-V analogous circuit
$F\to e$, $M\to L$, $B\to R$, $K\to 1/C$, velocity $\to$ mesh current. $B_1$ and $K_3$ are in series (same force), so in the F-V circuit they become the parallel pair $R = B_1 \parallel C = 1/K_3$, shared by mesh 1 ($M_1$) and mesh 2 ($M_2$); the loop inside the pair is mesh 3 (junction $x_3$).
```text
  L1  R=Bv1 C=1/K1      L2  R=Bv2+B2 C=1/K2
 +LLLL-/\/\--||---+-----LLLL---/\/\----||--+
 |                |                        |
(~) e       +-----+-----+                  |
 |    i1    |           |      i2          |
 |        R=B1   i3   C=1/K3               |
 |          |           |                  |
 |          +-----+-----+                  |
 |                |                        |
 +----------------+------------------------+
```
Mesh equations:
$$
\begin{aligned}
L_1\frac{di_1}{dt} + R_{v1}i_1 + \frac{1}{C_1}\int i_1dt + R_1(i_1 - i_3) &= e \\
R_1(i_3 - i_1) + \frac{1}{C_3}\int(i_3 - i_2)dt &= 0 \\
L_2\frac{di_2}{dt} + (R_{v2} + R_2)i_2 + \frac{1}{C_2}\int i_2dt + \frac{1}{C_3}\int(i_2 - i_3)dt &= 0
\end{aligned}
$$
($R_1 = B_1$ is shared by meshes 1 and 3; $C_3 = 1/K_3$ by meshes 3 and 2.)

## Check yourself

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

1. What is the equivalent of B1 in series with K3?
   A. B1 K3 s / (B1 s + K3)
   B. B1 s + K3
   C. B1 K3
   D. K3/B1
   Answer: A. Elements in series carry the same force, so their impedances combine like parallel resistors. (Key point: Series B1-K3)

2. Why does the transfer function have a zero at s = 0?
   A. The damper B1 transmits no static force
   B. K1 is zero
   C. M2 has no mass
   D. F is zero
   Answer: A. A steady force cannot move M2 because it must pass through B1, which gives no static force. (Key point: Zero at origin)

3. What is the order of the denominator?
   A. 3rd
   B. 4th
   C. 5th
   D. 6th
   Answer: C. The leading term is B1 M1 M2 s^5. (Key point: Order)

4. How do B1 and K3 appear in the F-V circuit?
   A. As a parallel pair R = B1 and C = 1/K3 shared by two meshes
   B. As a series pair in mesh 1
   C. As separate loads
   D. As sources
   Answer: A. Series mechanical elements become parallel electrical elements. (Key point: F-V circuit)

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