---
title: "Mass inside a mass: transfer function and F-I analogy"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-fi-m1-m2-nested"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fi-m1-m2-nested
site: BCT Engineering
---

# Mass inside a mass: transfer function and F-I analogy

Control Systems · Chapter 2: Component Modeling

- Asked in: 2078 Kartik, 6+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-fi-m1-m2-nested
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Consider displacement of mass M2 as output and find transfer function of the mechanical system as in figure. Also find F-I analogous circuit. [Figure: mass M1 is a hollow frame resting on the ground, with force f applied; mass M2 sits inside M1 and is connected to M1 through spring k1; M1 is connected to the right wall through spring k2]

## Short answer

Write equations for M₁ and M₂, eliminate X₁, yielding X₂(s)/F(s) = k₁ / [M₁M₂ s⁴ + (M₁k₁+M₂(k₁+k₂))s² + k₁k₂]. F–I analogy: f→i, M→C, k→1/L; circuit: C₁=M₁, C₂=M₂ to ground, L₂=1/k₂ from v₁ to ground, L₁=1/k₁ between v₁ and v₂.

## Answer

Let $x_1$ = displacement of the outer frame $M_1$ and $x_2$ = displacement of the inner mass $M_2$. Assumption: no friction (none is shown) between $M_1$ and ground or between $M_2$ and $M_1$.

#### Free body equations
- $M_1$: force $f$ acts; spring $k_2$ to the wall resists by $k_2x_1$; spring $k_1$ pulls by $k_1(x_1 - x_2)$.
- $M_2$: only spring $k_1$ acts, $k_1(x_2 - x_1)$.
$$
\begin{aligned}
M_1\ddot x_1 + k_1(x_1 - x_2) + k_2x_1 &= f \\
M_2\ddot x_2 + k_1(x_2 - x_1) &= 0
\end{aligned}
$$

#### Laplace form
$$
\begin{aligned}
(M_1s^2 + k_1 + k_2)X_1 - k_1X_2 &= F \\
-k_1X_1 + (M_2s^2 + k_1)X_2 &= 0
\end{aligned}
$$
From the second equation, $X_1 = \dfrac{M_2s^2 + k_1}{k_1}X_2$. Substituting:
$$
\left[(M_1s^2 + k_1 + k_2)(M_2s^2 + k_1) - k_1^2\right]X_2 = k_1F
$$

#### Transfer function
$$
\frac{X_2(s)}{F(s)} = \frac{k_1}{M_1M_2s^4 + \left[M_1k_1 + M_2(k_1 + k_2)\right]s^2 + k_1k_2}
$$
With no damping the poles lie on the $j\omega$ axis (undamped oscillation at two natural frequencies).

#### F-I analogous circuit
$f\to i$, $M\to C$, $k\to 1/L$, velocity $\to$ node voltage. Note that $M_2$ is referred to the ground reference (absolute velocity), so its capacitor goes to ground.
```text
 v1           L1    v2
 +----+----+--LLLL--+
 |    |    |        |
(i)   C1   L2       C2
 |    |    |        |
 +----+----+--------+ gnd
```
($C_1 = M_1$, $L_2 = 1/k_2$, $L_1 = 1/k_1$, $C_2 = M_2$.)
$$
\begin{aligned}
C_1\frac{dv_1}{dt} + \frac{1}{L_2}\int v_1dt + \frac{1}{L_1}\int(v_1 - v_2)dt &= i \\
C_2\frac{dv_2}{dt} + \frac{1}{L_1}\int(v_2 - v_1)dt &= 0
\end{aligned}
$$

## Check yourself

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

1. Which equation describes the inner mass M2?
   A. M2 x2'' + k1 x2 = f
   B. M2 x2'' + (k1 + k2)(x2 - x1) = 0
   C. M2 x2'' + k2 x2 = f
   D. M2 x2'' + k1(x2 - x1) = 0
   Answer: D. Only spring k1 acts on M2, with force k1 times the relative displacement. (Key point: Equations of motion)

2. What is the constant term of the denominator of X2/F?
   A. k1^2
   B. k1 + k2
   C. k1 k2
   D. M1 M2
   Answer: C. Setting s = 0 in (k1+k2)k1 - k1^2 gives k1 k2. (Key point: Transfer function)

3. In the F-I circuit, where is the capacitor C2 = M2 connected?
   A. Between node 1 and node 2
   B. From node v2 to ground, because M2 is referred to the ground reference
   C. In series with the source
   D. In parallel with L2
   Answer: B. Mass is referenced to ground (absolute velocity), so its capacitor goes to ground. (Key point: F-I circuit)

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