---
title: "Two-mass spring system X1/F and F-V analogy"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-fv-m1-m2-k1-k2"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-m1-m2-k1-k2
site: BCT Engineering
---

# Two-mass spring system X1/F and F-V analogy

Control Systems · Chapter 2: Component Modeling

- Asked in: 2067 Asar (old course), 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-fv-m1-m2-k1-k2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Find the transfer function X1(s)/F(s) of the mechanical system shown in figure 1. Also find the force-voltage analogy of the same system. [Figure 1: wall – spring K1 – mass M1 (displacement x1(t)) – spring K2 – mass M2 (displacement x2(t), force F(t) applied); ground friction B1 = B2 = 0]

## Short answer

Equations: M1x1''+K1x1+K2(x1-x2)=0, M2x2''+K2(x2-x1)=F. Transfer function X1(s)/F(s)=K2/[M1M2s⁴+(M1K2+M2(K1+K2))s²+K1K2]. Force-voltage analogy: L=M, C=1/K, e=F, mesh currents as velocities.

## Answer

Displacements $x_1$ ($M_1$) and $x_2$ ($M_2$, where $F$ acts). No friction ($B_1 = B_2 = 0$).

#### Equations of motion
$$
\begin{aligned}
M_1\ddot x_1 + K_1x_1 + K_2(x_1 - x_2) &= 0 \\
M_2\ddot x_2 + K_2(x_2 - x_1) &= F(t)
\end{aligned}
$$

#### Laplace form
$$
\begin{aligned}
(M_1s^2 + K_1 + K_2)X_1 - K_2X_2 &= 0 \\
-K_2X_1 + (M_2s^2 + K_2)X_2 &= F
\end{aligned}
$$
From the first equation, $X_2 = \dfrac{M_1s^2 + K_1 + K_2}{K_2}X_1$. Substituting into the second:
$$
\left[\frac{(M_1s^2 + K_1 + K_2)(M_2s^2 + K_2)}{K_2} - K_2\right]X_1 = F
$$

#### Transfer function
$$
\frac{X_1(s)}{F(s)} = \frac{K_2}{(M_1s^2 + K_1 + K_2)(M_2s^2 + K_2) - K_2^2}
$$
$$
\frac{X_1(s)}{F(s)} = \frac{K_2}{M_1M_2s^4 + \left[M_1K_2 + M_2(K_1 + K_2)\right]s^2 + K_1K_2}
$$
Only even powers of $s$ appear because there is no damping; the poles are on the $j\omega$ axis (two undamped natural frequencies).

#### Force-voltage analogy
$F\to e$, $M\to L$, $K\to 1/C$, velocity $\to$ current.
$$
\begin{aligned}
L_1\frac{di_1}{dt} + \frac{1}{C_1}\int i_1dt + \frac{1}{C_2}\int(i_1 - i_2)dt &= 0 \\
L_2\frac{di_2}{dt} + \frac{1}{C_2}\int(i_2 - i_1)dt &= e
\end{aligned}
$$
with $L_1 = M_1$, $L_2 = M_2$, $C_1 = 1/K_1$, $C_2 = 1/K_2$.
```text
    L1 = M1                 L2 = M2
 +---LLLL----+-------------LLLL----+
 |           |                     |
C1=1/K1    C2=1/K2                (~) e
 |    i1     |         i2          |
 +-----------+---------------------+
```

## Check yourself

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

1. Which equation describes M1?
   A. M1 x1'' + K1 x1 + K2(x1 - x2) = 0
   B. M1 x1'' + K1 x1 = F
   C. M1 x1'' + K2(x1 - x2) = F
   D. M1 x1'' + K1 x1 + K2 x2 = 0
   Answer: A. M1 is held by K1 to the wall and by K2 to M2, and F acts on M2, not M1. (Key point: Equations of motion)

2. What is the numerator of X1(s)/F(s)?
   A. M2 s^2 + K2
   B. K2
   C. K1 K2
   D. M1 s^2 + K1 + K2
   Answer: B. F reaches M1 only through the coupling spring K2. (Key point: Transfer function numerator)

3. Why are only even powers of s in the denominator?
   A. There is no damping
   B. Springs have no dynamics
   C. F is constant
   D. M1 = M2
   Answer: A. Odd powers come from damper terms, which are zero, so the poles lie on the jω axis. (Key point: Even powers)

4. In the F-V analogous circuit, C1 and C2 equal
   A. K1 and K2
   B. 1/K1 and 1/K2
   C. M1 and M2
   D. 1/M1 and 1/M2
   Answer: B. Springs map to capacitors with C = 1/K, and masses map to inductors L = M. (Key point: F-V analogy)

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