Why does this mechanical system have 4 state variables?

Each mass is second order, so a displacement and a velocity are needed per mass

There are four elements: two springs and two dampers

There are two outputs and two inputs

Each spring and damper pair adds one state

Which is the correct equation of motion for m1?

m1 x1'' = F - k1(x1 - x2) - b1(x1' - x2')

m1 x1'' = F + k1(x1 - x2) + b1(x1' - x2')

m1 x1'' = -k1(x1 - x2) - b1(x1' - x2')

Which forces act on m2?

k1, b1 pull it by the relative motion with m1, while k2, b2 hold it back against the wall

2080 Baisakh · 8 marks

State space model of two-mass system

Short answer

Use states x1, ẋ1, x2, ẋ2. The state-space model is ẋ=Ax+BF, y=Cx with A = [[0,1,0,0],[-k1/m1,-b1/m1,k1/m1,b1/m1],[0,0,0,1],[k1/m2,b1/m2,-(k1+k2)/m2,-(b1+b2)/m2]], B = [0,1/m1,0,0]ᵀ, C = [[1,0,0,0],[0,0,1,0]], D=0.

Obtain a state-space representation of the mechanical system shown in the figure where external force F is the input and the displacements of the masses x1 and x2 are the outputs. [Figure: wall – spring k2 parallel with damper b2 – mass m2 (displacement x2, on rollers) – spring k1 parallel with damper b1 – mass m1 (displacement x1, on rollers, force F applied)]

A state-space model is obtained by writing Newton's second law for each mass and choosing the displacements and velocities as state variables (one pair per mass, so 4 states).

System and assumptions

 wall |--[k2 || b2]--[ m2 ]--[k1 || b1]--[ m1 ]--> F
                      x2 ->               x1 ->

Masses run on rollers, so there is no friction with the ground.
$x_1$ , $x_2$ are measured from equilibrium, positive to the right.
$k_1$ , $b_1$ connect $m_1$ and $m_2$ ; $k_2$ , $b_2$ connect $m_2$ to the wall.

Equations of motion

Free body of $m_1$ (force $F$ , spring and damper $k_1$ , $b_1$ resisting relative motion):

m_1\ddot{x}_1 = F - k_1(x_1 - x_2) - b_1(\dot{x}_1 - \dot{x}_2)

Free body of $m_2$ (pulled by $k_1$ , $b_1$ ; held back by $k_2$ , $b_2$ ):

m_2\ddot{x}_2 = k_1(x_1 - x_2) + b_1(\dot{x}_1 - \dot{x}_2) - k_2x_2 - b_2\dot{x}_2

Choice of state variables

z_1 = x_1,\quad z_2 = \dot{x}_1,\quad z_3 = x_2,\quad z_4 = \dot{x}_2,\quad u = F

State equations

\begin{aligned} \dot{z}_1 &= z_2 \\ \dot{z}_2 &= -\frac{k_1}{m_1}z_1 - \frac{b_1}{m_1}z_2 + \frac{k_1}{m_1}z_3 + \frac{b_1}{m_1}z_4 + \frac{1}{m_1}u \\ \dot{z}_3 &= z_4 \\ \dot{z}_4 &= \frac{k_1}{m_2}z_1 + \frac{b_1}{m_2}z_2 - \frac{k_1 + k_2}{m_2}z_3 - \frac{b_1 + b_2}{m_2}z_4 \end{aligned}

Matrix form

\begin{bmatrix}\dot{z}_1\\ \dot{z}_2\\ \dot{z}_3\\ \dot{z}_4\end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0\\ -\frac{k_1}{m_1} & -\frac{b_1}{m_1} & \frac{k_1}{m_1} & \frac{b_1}{m_1}\\ 0 & 0 & 0 & 1\\ \frac{k_1}{m_2} & \frac{b_1}{m_2} & -\frac{k_1+k_2}{m_2} & -\frac{b_1+b_2}{m_2} \end{bmatrix} \begin{bmatrix}z_1\\ z_2\\ z_3\\ z_4\end{bmatrix} + \begin{bmatrix}0\\ \frac{1}{m_1}\\ 0\\ 0\end{bmatrix}F

Output equation

The outputs are the two displacements $y_1 = x_1$ and $y_2 = x_2$ :

\begin{bmatrix}y_1\\ y_2\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\end{bmatrix} \begin{bmatrix}z_1\\ z_2\\ z_3\\ z_4\end{bmatrix} + \begin{bmatrix}0\\ 0\end{bmatrix}F

Summary

Matrix	Size	Meaning
$A$	$4\times 4$	Masses, springs, dampers
$B$	$4\times 1$	Force acts on $m_1$ only
$C$	$2\times 4$	Picks $x_1$ and $x_2$
$D$	$2\times 1$	Zero (no direct path)

Answer: $\dot{z} = Az + BF$ , $y = Cz$ with $A$ , $B$ , $C$ as above and $D = 0$ . Physical check: in the static case ( $\dot{z} = 0$ ), the equations give $k_2x_2 = F$ and $k_1(x_1 - x_2) = F$ , i.e. the two springs carry the same force in series, as expected.

✓ Answer checked by Claude Opus 5.5 (high effort)

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