Training Multiphysics Differential Equations Coupled Mass–Spring–Damper Systems
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Coupled Mass–Spring–Damper Systems

30 min Multiphysics Differential Equations

Coupled Mass–Spring–Damper Systems

Multiphysics problems arise when two or more physical domains interact through shared state variables. The simplest case — and the one most useful for building intuition — is a chain of masses connected by springs and dashpots. Each mass obeys Newton's second law, but its motion depends on its neighbors through spring forces, producing a coupled system of second-order ODEs.

The governing equations are $m_1 \ddot{x}_1 + c_1 \dot{x}_1 + k_1 x_1 + k_2(x_1 - x_2) = F_1(t)$ and a similar equation for $m_2$. When written in matrix form $M\ddot{\mathbf{x}} + C\dot{\mathbf{x}} + K\mathbf{x} = \mathbf{F}(t)$, the eigenvalue problem $\det(K - \omega^2 M) = 0$ gives the natural frequencies, and the eigenvectors give the mode shapes.

Real structures — buildings, aircraft, vehicles, and MEMS resonators — are routinely reduced to such coupled systems. In this lesson we derive the equations, compute natural frequencies for a two-mass chain, and look at forced response.

Two-Mass Coupled System

$$m_1 \ddot{x}_1 = -k_1 x_1 - k_2(x_1 - x_2) - c_1 \dot{x}_1 + F_1(t)$$$$m_2 \ddot{x}_2 = -k_2(x_2 - x_1) - c_2 \dot{x}_2 + F_2(t)$$

Natural Frequencies (Undamped)

For equal masses $m$ and equal springs $k$, the two modes are $\omega_1 = \sqrt{k/m}$ (in-phase) and $\omega_2 = \sqrt{3k/m}$ (out-of-phase).

Example 1 — Two-mass in-phase mode

Find $\omega_1$ for $m = 2\,\text{kg}$ and $k = 800\,\text{N/m}$.

$\omega_1 = \sqrt{800/2} = \sqrt{400} = 20\,\text{rad/s}$, so $f_1 = \omega_1/(2\pi) \approx 3.18\,\text{Hz}$.

Example 2 — Out-of-phase mode

Same $m$ and $k$ as Example 1: find $\omega_2$.

$\omega_2 = \sqrt{3 \cdot 800 / 2} = \sqrt{1200} \approx 34.64\,\text{rad/s}$, i.e. $f_2 \approx 5.51\,\text{Hz}$.

Example 3 — Unequal stiffness

If $m_1 = m_2 = 1\,\text{kg}$, $k_1 = 100\,\text{N/m}$, $k_2 = 200\,\text{N/m}$, find both natural frequencies from $\det(K - \omega^2 M) = 0$.

$K = \begin{bmatrix}300 & -200\\-200 & 200\end{bmatrix}$, $M = I$. The characteristic equation is $(300-\lambda)(200-\lambda) - 40000 = 0 \Rightarrow \lambda^2 - 500\lambda + 20000 = 0$.

$\lambda = (500 \pm \sqrt{250000 - 80000})/2 = (500 \pm 412.3)/2$, giving $\lambda_1 = 43.84$, $\lambda_2 = 456.16$.

So $\omega_1 \approx 6.62\,\text{rad/s}$, $\omega_2 \approx 21.36\,\text{rad/s}$.

Interactive Demo: Two-Mass Coupled Oscillator
ω₁ =6.62rad/s
ω₂ =21.36rad/s
f₁ =1.05Hz
f₂ =3.40Hz

Practice Problems

1. For $m=m_1=m_2=0.5$ kg and $k_1=k_2=500$ N/m, compute both natural frequencies.
2. What physical meaning does the in-phase mode have?
3. Write the matrix form $M\ddot{\mathbf{x}} + K\mathbf{x} = 0$ for the example in text.
4. Why does adding a damper $c$ not change the eigenvectors for light damping?
5. If $k_2 \to 0$, what happens to the two modes?
6. Name two engineering applications of coupled-mass analysis.
Show Answer Key

1. $\omega_1 = \sqrt{500/0.5}=31.62$ rad/s, $\omega_2 = \sqrt{1500/0.5}=54.77$ rad/s.

2. Both masses oscillate together in the same direction; the middle spring is barely compressed.

3. $\begin{bmatrix}m&0\\0&m\end{bmatrix}\ddot{\mathbf{x}} + \begin{bmatrix}k_1+k_2&-k_2\\-k_2&k_2\end{bmatrix}\mathbf{x} = 0$.

4. Small $C$ acts as a perturbation that changes decay rates more than it changes mode shapes.

5. The system decouples into two independent single-mass oscillators $\omega=\sqrt{k_1/m_1}$ and undamped $m_2$ free.

6. Vehicle suspension tuning, MEMS resonators, aircraft wing flutter, tuned mass dampers in skyscrapers.

RLC Circuits & Electromechanical Coupling