One pattern, many machines

A block, a spinning shaft, and an electrical circuit all obey the very same equation. Learn the pattern once, and you can read them all.

Source: course notes, Week 1 foundations.

Before you start

What you need first

The mass–spring–damper — $m\ddot x + c\dot x + kx = F(t)$ and its natural frequency $\omega_n=\sqrt{k/m}$ (previous topic).

What you'll be able to do

Recognise the universal second-order pattern.
Map a shaft and an RLC circuit onto it.
Read off a natural frequency by analogy.

The pattern behind them all

$$(\text{inertia})\,\ddot{(\ )} + (\text{damping})\,\dot{(\ )} + (\text{stiffness})\,(\ ) = (\text{input})$$

The mass–spring–damper was one example: $m\ddot x + c\dot x + kx = F(t)$. But the shape — an inertia term, a damping term, a restoring term, and a drive — shows up far beyond a block on a spring.

Learn to read this one pattern and you can read a circuit, a shaft, and a robot joint with the same eyes. Understanding the pattern beats memorising machines.

The same equation — a spinning shaft

$$J\,\ddot\theta + b\,\dot\theta + k\,\theta = T$$

where:
Symbol	Meaning	SI unit
J	rotational inertia (the "mass")	kg·m²
b	rotational damping	N·m·s
k	torsional stiffness	N·m/rad
$\theta$	twist angle	rad
T	applied torque (the input)	N·m

Swap "sliding" for "spinning" and the very same equation appears.

The same equation — an R-L-C circuit

$$L\,\ddot q + R\,\dot q + \tfrac{1}{C}\,q = V$$

where:
Symbol	Meaning	SI unit
L	inductance (the "mass")	H
R	resistance (the "damping")	Ω
$1/C$	inverse capacitance (the "stiffness")	1/F
q	charge	C
V	applied voltage (the input)	V

An electrical and a mechanical engineer are secretly solving the same equation.

One pattern, many machines

Role	Spring–mass	Shaft	R-L-C circuit
inertia	$m$	$J$	$L$
damping	$c$	$b$	$R$
stiffness	$k$	$k$	$1/C$
input	$F$	$T$	$V$
natural freq.	$\sqrt{k/m}$	$\sqrt{k/J}$	$\sqrt{1/(LC)}$

🔭 Looking ahead: a robot joint will show the same characters — an inertia term, a damping term, and a gravity (restoring) term. Same pattern, bigger machine.

📐 Worked example

Natural frequency of a shaft — by analogy

A torsional shaft has rotational inertia $J = 0.02\ \text{kg·m}^2$ and torsional stiffness $k = 8\ \text{N·m/rad}$. Find its natural frequency.

1The shaft matches the spring pattern, so $\omega_n=\sqrt{k/J}$ (the rotational twin of $\sqrt{k/m}$):

$$\omega_n=\sqrt{\frac{k}{J}}=\sqrt{\frac{8}{0.02}}=\sqrt{400}$$

2Take the square root:

$$\omega_n = 20\ \text{rad/s}$$

We didn't re-derive anything — we recognised the pattern and reused the spring's result with $m\to J$. That's the payoff of one pattern, many machines.

✏️ Try it yourself

An R-L-C circuit has $L = 2\ \text{H}$ and $1/C = 200\ \text{F}^{-1}$. Find its natural frequency by analogy with the spring ($\omega_n=\sqrt{\text{stiffness}/\text{inertia}}$).

Step 1. Stiffness $=1/C=200$, inertia $=L=2$, so $\omega_n=\sqrt{\dfrac{1/C}{L}}=\sqrt{\dfrac{200}{2}}=\sqrt{100}$. Step 2. $\omega_n = 10\ \text{rad/s}$.

Recap — the whole topic on one screen

Idea	What you own now
The universal pattern	inertia·$\ddot{(\ )}$ + damping·$\dot{(\ )}$ + stiffness·$(\ )$ = input
Shaft	$J\ddot\theta + b\dot\theta + k\theta = T$
R-L-C circuit	$L\ddot q + R\dot q + \tfrac{1}{C}q = V$
Natural frequency	Always $\sqrt{\text{stiffness}/\text{inertia}}$

Next topic

The pendulum by Newton

Next we run the Newton recipe on a turning system — the pendulum — and meet our first nonlinear equation of motion.

→ The pendulum by Newton

Role	Spring–mass	Shaft	R-L-C circuit
inertia	\(m\)	\(J\)	\(L\)
damping	\(c\)	\(b\)	\(R\)
stiffness	\(k\)	\(k\)	\(1/C\)
input	\(F\)	\(T\)	\(V\)
natural freq.	\(\sqrt{k/m}\)	\(\sqrt{k/J}\)	\(\sqrt{1/(LC)}\)

Idea	What you own now
The universal pattern	inertia·\(\ddot{(\ )}\) + damping·\(\dot{(\ )}\) + stiffness·\((\ )\) = input
Shaft	\(J\ddot\theta + b\dot\theta + k\theta = T\)
R-L-C circuit	\(L\ddot q + R\dot q + \tfrac{1}{C}q = V\)
Natural frequency	Always \(\sqrt{\text{stiffness}/\text{inertia}}\)