---
title: "Nested-mass system transfer function with F-V and F-I"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-fv-fi-m1-m2-nested"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-fi-m1-m2-nested
site: BCT Engineering
---

# Nested-mass system transfer function with F-V and F-I

Control Systems · Chapter 2: Component Modeling

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

## Question

Find transfer function and develop F-V and F-I analogous circuits of the following figure. [Figure: mass M2 is a hollow frame resting on the ground (viscous friction B2 between M2 and ground), connected to the left wall through spring K2; mass M1 is a plate sliding inside M2 with viscous friction B1 between M1 and M2; force F(t) pulls M1 through spring K1]

## Short answer

Force acts on M1 directly. Coupled equations give X2(s)/F(s) = B1/Δ', with Δ' = M1M2s^3+[ B1(M1+M2)+B2M1]s^2+(B1B2+K2M1)s+B1K2. F-V circuit: three loops; K1 as capacitor between loop0 and loop1, B1 as resistor between loop1 and loop2. F-I analogue: current source, nodes; K1→L=1/K1, B1→R=1/B1.

## Answer

Displacements: $x_1$ (plate $M_1$), $x_2$ (frame $M_2$). The force $F$ acts on the free end of spring $K_1$; since that end has no mass, the spring transmits the full force, $K_1(x_0 - x_1) = F$, so $F$ effectively acts on $M_1$. (Here $x_0$ is the free-end displacement.)

#### Equations of motion
- $M_1$: pulled by $F$; viscous friction $B_1$ relative to $M_2$.
- $M_2$: friction $B_1$ from $M_1$, ground friction $B_2$, spring $K_2$ to the wall.
$$
\begin{aligned}
M_1\ddot x_1 + B_1(\dot x_1 - \dot x_2) &= F \\
M_2\ddot x_2 + B_1(\dot x_2 - \dot x_1) + B_2\dot x_2 + K_2x_2 &= 0
\end{aligned}
$$
Laplace form:
$$
\begin{aligned}
(M_1s^2 + B_1s)X_1 - B_1sX_2 &= F \\
-B_1sX_1 + \left[M_2s^2 + (B_1 + B_2)s + K_2\right]X_2 &= 0
\end{aligned}
$$

#### Transfer functions
Determinant (after cancelling a common $s$):
$$
\Delta' = (M_1s + B_1)\left[M_2s^2 + (B_1 + B_2)s + K_2\right] - B_1^2s = M_1M_2s^3 + \left[B_1(M_1 + M_2) + B_2M_1\right]s^2 + (B_1B_2 + K_2M_1)s + B_1K_2
$$
$$
\frac{X_2(s)}{F(s)} = \frac{B_1}{\Delta'}
$$
$$
\frac{X_1(s)}{F(s)} = \frac{M_2s^2 + (B_1 + B_2)s + K_2}{s\,\Delta'}
$$
and for the point where the force is applied, $X_0(s) = X_1(s) + F(s)/K_1$.

#### F-V analogous circuit
$F\to e$, $M\to L$, $B\to R$, $K\to 1/C$. Mesh $i_0$ = free-end velocity, $i_1$ = velocity of $M_1$, $i_2$ = velocity of $M_2$. $K_1$ is shared by meshes 0 and 1; $B_1$ (relative friction) by meshes 1 and 2.
```text
              L1=M1             L2=M2  R2=B2  C2=1/K2
 +------+-----LLLL----+---------LLLL---/\/\----||--+
 |      |             |                            |
(~) e  C=1/K1       R1=B1                          |
 |  i0  |      i1     |            i2              |
 +------+-------------+----------------------------+
```

#### F-I analogous circuit
$F\to i$, $M\to C$, $B\to 1/R$, $K\to 1/L$. Node 0 = free end, node 1 = $M_1$, node 2 = $M_2$; $C_1 = M_1$, $C_2 = M_2$, $R_2 = 1/B_2$, $L_2 = 1/K_2$.
```text
  v0    L=1/K1    v1   R1=1/B1    v2
 +---+---LLLL---+---+---/\/\/---+---+---+---+
 |                  |               |   |   |
(i)                C1              C2  R2  L2
 |                  |               |   |   |
 +------------------+---------------+---+---+ gnd
```

## Check yourself

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

1. Why can the force F be treated as acting directly on M1?
   A. The free end of K1 has no mass, so the spring transmits the full force
   B. K1 is infinitely stiff
   C. M2 absorbs the force
   D. F is cancelled by B1
   Answer: A. A massless end has zero net force, so K1(x0 - x1) = F and the spring passes F on to M1. (Key point: Massless spring end)

2. How does the friction B1 between plate M1 and frame M2 appear in the equations?
   A. As B1(x1' - x2') in M1's equation and B1(x2' - x1') in M2's equation
   B. As B1 x1' only in M1's equation
   C. As B1 x2' in both equations
   D. As B1(x1 - x2)
   Answer: A. B1 depends on relative velocity, so it appears with opposite signs on the two masses. (Key point: Relative friction B1)

3. What is X2(s)/F(s) for the nested system?
   A. B1/Δ', with Δ' a cubic polynomial
   B. B1 s/Δ'
   C. K2/Δ'
   D. (M1 s + B1)/Δ'
   Answer: A. Solving the Laplace equations (after cancelling a common s) gives X2/F = B1/Δ'. (Key point: Transfer function X2/F)

4. In the F-V analogy used here, how are the quantities mapped?
   A. F to e, M to L, B to R, K to 1/C
   B. F to i, M to C, B to 1/R, K to 1/L
   C. F to e, M to C, B to R, K to L
   D. F to i, M to L, B to R, K to C
   Answer: A. In the force-voltage analogy force becomes voltage, mass becomes inductance, friction becomes resistance and spring constant becomes 1/C. (Key point: F-V analogy mapping)

## Related in Control Systems

- [Two-mass system transfer function and F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-two-mass-k1k2b1b2) — Control Systems, chapter 2
- [Mass inside a mass: transfer function and F-I analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fi-m1-m2-nested) — Control Systems, chapter 2
- [Single-mass system transfer function with F-V and F-I](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-fi-k1-m2-k2b2) — Control Systems, chapter 2
- [Velocity transfer function of two-mass system with analogies](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-velocity-m2-fv-fi) — Control Systems, chapter 2
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