ME3120 · Dynamic Modelling

Theme 2 · Newton's method

The mass–spring–damper

The single most important equation in all of dynamics — built from scratch, one term at a time, and read like a sentence.

Source: course notes, after a standard 1-DOF oscillator treatment.

Before you start

What you need first

Newton's laws & free-body diagrams — $\sum F = m\ddot x$, and how to draw the forces on one body.
Generalized coordinates — choosing one coordinate $x$ for a 1-DOF system.

What you'll be able to do

Build the equation of motion of a spring–mass system with Newton.
Find its natural frequency $\omega_n$.
Add damping and a drive, and name what every term does.

Step 1 · the picture

The system: a block on a spring

A block of mass $m$ sits on a frictionless table, tied to a wall by a spring of stiffness $k$.

The system is just the block. We let $x$ be the distance from the spring's rest position, and pick positive $x$ to the right.

One degree of freedom → one coordinate $x$. That single number says exactly where the block is.

Block on a frictionless table, held by a spring of stiffness $k$.

Step 3 · the force

The spring force (Hooke's law)

The only horizontal force is the spring. Hooke's law says the force is proportional to the stretch, and points back toward rest:

$$F_{\text{spring}} = -k\,x$$

The minus sign means the spring always pulls back toward rest. Pulled right ($x>0$) → it pulls left. Pushed left ($x<0$) → it pushes right.

$k$ is the stiffness (N/m): a bigger $k$ means a stronger pull-back for the same stretch.

Step 3 · isolate the body

The free-body diagram

Block displaced right by $x$; the spring force points left.

Isolate the block and draw every outside force. On a frictionless table, in the horizontal direction, there is only one:

Only one horizontal force acts — the spring, $F=-kx$. That single force is our $\sum F$.

(Gravity and the table's normal force act vertically and cancel, so they play no part in the horizontal motion.)

Step 4 · derive

Derive the equation of motion

1Start from Newton's law along $x$:

$$\sum F = m\,\ddot x$$

2Put in the only force, the spring $F=-kx$:

$$-k\,x = m\,\ddot x$$

3Add $kx$ to both sides to gather terms on the left:

$$m\,\ddot x + k\,x = 0$$

The equation of motion

$$m\,\ddot x + k\,x = 0$$

where:
Symbol	Meaning	SI unit
m	mass of the block	kg
$\ddot x$	acceleration of the block	m/s²
k	spring stiffness	N/m
x	displacement from rest	m

This single line is the model of the spring–mass system.

The $m\ddot x$ term is inertia (it resists changes in speed); the $kx$ term is stiffness (it pulls back toward rest). They fight each other → the block oscillates back and forth.

Natural frequency

Any free oscillation can be written in one standard form, $\ \ddot x + \omega_n^{2}\,x = 0$, where $\omega_n$ is its natural frequency. We bend our equation into this shape, then read $\omega_n$ off.

1Divide the whole equation by $m$:

$$\ddot x + \frac{k}{m}\,x = 0$$

2Compare with $\ \ddot x + \omega_n^{2} x = 0$, so $\omega_n^{2}=k/m$. Take the square root:

$$\omega_n = \sqrt{\dfrac{k}{m}}\,$$natural frequency (rad/s)

Units check: $\left[\dfrac{k}{m}\right]=\dfrac{\text{N/m}}{\text{kg}}=\dfrac{1}{\text{s}^2}$, so $\omega_n$ comes out in $1/\text{s}$ ✓. Stiffer spring → faster wobble; heavier mass → slower wobble.

📐 Worked example

Find the natural frequency and period

A block of mass $m = 2\ \text{kg}$ is held by a spring of stiffness $k = 200\ \text{N/m}$ on a frictionless table. Find the natural frequency $\omega_n$ and the period $T$ of its free wobble.

1Natural frequency from $\omega_n=\sqrt{k/m}$:

$$\omega_n=\sqrt{\frac{200}{2}}=\sqrt{100}=10\ \text{rad/s}$$

2Period from $T = 2\pi/\omega_n$:

$$T=\frac{2\pi}{\omega_n}=\frac{2\pi}{10}\approx 0.63\ \text{s}$$

Read it back: the block completes one full back-and-forth every $0.63$ seconds. If we made the block four times heavier, $\omega_n$ would halve (to $5$ rad/s) and the period would double — heavier means slower.

Make it real: add damping and a push

A damper (like thick oil) gives a force that fights motion, sized by speed: $F_{\text{damper}}=-c\,\dot x$. An outside push $F(t)$ goes on the right. Putting all three together:

Mass–spring–damper: spring $k$ and damper $c$ in parallel, driven by $F(t)$.

$$m\,\ddot x + c\,\dot x + k\,x = F(t)$$

the new terms:
Symbol	Meaning	SI unit
c	damping coefficient	N·s/m
$\dot x$	velocity of the block	m/s
$F(t)$	applied (driving) force	N

This is the famous mass–spring–damper equation — the most important equation in all of dynamics.

Read every term

Term	Name	What it physically does
$m\ddot x$	inertia	resists changes in speed (heaviness)
$c\dot x$	damping	removes energy; makes the wobble die down
$kx$	stiffness	pulls back toward rest
$F(t)$	input	the outside push driving the system

🔭 Looking ahead: when we model a robot joint, we will meet exactly these characters — an inertia term, a damping term, and a restoring/gravity term. Learn to read them here and you can read them on any machine.

✏️ Try it yourself

A block of mass $m = 0.5\ \text{kg}$ is held by a spring of stiffness $k = 32\ \text{N/m}$. Find its natural frequency $\omega_n$ and period $T$.

Step 1. $\omega_n=\sqrt{k/m}=\sqrt{32/0.5}=\sqrt{64}=8\ \text{rad/s}$ Step 2. $T=\dfrac{2\pi}{\omega_n}=\dfrac{2\pi}{8}\approx 0.79\ \text{s}$ Read it: a lighter block on a softer spring — it wobbles a touch slower than the worked example (period $0.79$ s vs $0.63$ s).

Common mistakes to avoid

Mistake	Fix
Dropping the spring's minus sign	The spring force is $-kx$: always toward rest.
Putting an "$m\ddot x$ arrow" on the FBD	Only outside forces go on the diagram; $m\ddot x$ lives on the other side of the equation.
Forgetting to divide by $m$ before reading $\omega_n$	The standard form is $\ddot x + \omega_n^2 x = 0$ — the $\ddot x$ must stand alone.
Mixing up $c$ and $k$ roles	$c$ multiplies velocity $\dot x$; $k$ multiplies position $x$.

Recap — the whole topic on one screen

$$m\,\ddot x + c\,\dot x + k\,x = F(t)\qquad\qquad \omega_n=\sqrt{\tfrac{k}{m}}$$

Idea	What you own now
Build the model	Newton + one spring force → $m\ddot x + kx = 0$
Natural frequency	Bend into standard form, read $\omega_n=\sqrt{k/m}$
The full equation	Add damping $c\dot x$ and a drive $F(t)$
Read it	Name what inertia, damping, stiffness, and input each do

Next topic

One pattern, many machines

The very same equation describes a spinning shaft and an electrical R-L-C circuit. Next we'll see why a block, a shaft, and a circuit are all secretly solving $(\text{inertia})\,\ddot{(\,)} + (\text{damping})\,\dot{(\,)} + (\text{stiffness})\,(\,) = (\text{input})$.

→ One pattern, many machines

Mistake	Fix
Dropping the spring's minus sign	The spring force is \(-kx\): always toward rest.
Putting an "\(m\ddot x\) arrow" on the FBD	Only outside forces go on the diagram; \(m\ddot x\) lives on the other side of the equation.
Forgetting to divide by \(m\) before reading \(\omega_n\)	The standard form is \(\ddot x + \omega_n^2 x = 0\) — the \(\ddot x\) must stand alone.
Mixing up \(c\) and \(k\) roles	\(c\) multiplies velocity \(\dot x\); \(k\) multiplies position \(x\).

Symbol	Meaning	SI unit
m	mass of the block	kg
\(\ddot x\)	acceleration of the block	m/s²
k	spring stiffness	N/m
x	displacement from rest	m

Symbol	Meaning	SI unit
c	damping coefficient	N·s/m
\(\dot x\)	velocity of the block	m/s
\(F(t)\)	applied (driving) force	N

Term	Name	What it physically does
\(m\ddot x\)	inertia	resists changes in speed (heaviness)
\(c\dot x\)	damping	removes energy; makes the wobble die down
\(kx\)	stiffness	pulls back toward rest
\(F(t)\)	input	the outside push driving the system

Idea	What you own now
Build the model	Newton + one spring force → \(m\ddot x + kx = 0\)
Natural frequency	Bend into standard form, read \(\omega_n=\sqrt{k/m}\)
The full equation	Add damping \(c\dot x\) and a drive \(F(t)\)
Read it	Name what inertia, damping, stiffness, and input each do