Hooke's Law — Physica

Equivalent forms

F_s = -kx

\vec{F} = -k\vec{x}

The linear approximation that underlies nearly all small-oscillation physics.

Symbol	Name	SI unit	Dimension	Range
$F$	Restoring Forceoutput Force exerted by the spring (magnitude)	N	MLT^-2	0 – 500
$k$	Spring Constant Stiffness of the spring	N/m	MT^-2	1 – 500
$x$	Displacement Displacement from the equilibrium position	m	L	0 – 2

Unit systems

Symbol	SI	CGS	Imperial
$F$	N	dyn	lbf
$k$	N/m	dyn/cm	lbf/in
$x$	m	cm	in

Where it holds

Valid within the elastic limit of the material. Breaks down for large deformations where stress-strain is nonlinear.

Dimensional analysis

F \to [MLT^{-2}]

k\cdot x \to [MT^{-2}]\cdot [L] = [MLT^{-2}]

Both sides are

[MLT^{-2}]

= Newton.

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Discovery

Robert Hooke · 1676

First published as a Latin anagram 'ceiiinosssttuv' (ut tensio, sic vis — as the extension, so the force).

Try this

A bungee cord stretches 4 m under a 70 kg jumper. What's the cord's spring constant?

Find the spring constant k given that a 70 kg person stretches the cord 4 m at equilibrium, where the restoring force equals weight.

Research status: stable

Real-world applications

Vehicle suspension systems
Seismograph design
Atomic bond modeling
Mattress and cushion engineering

Common misconceptions

Hooke's law is not a fundamental law — it's a linear approximation
The negative sign indicates direction (restoring), not that force is negative in magnitude
Not all springs obey Hooke's law — only within the elastic limit

Experimental verification

Verified with calibrated springs, atomic force microscopes, and crystal lattice vibration measurements.

Derivation

Empirical law: for small deformations, the restoring force is proportional to displacement.

Derived from Taylor expansion of any potential energy minimum:

U(x) \approx U_{0} + \tfrac{1}{2}kx^{2}

.

Limiting cases

x \to 0

⟶

F \to 0

No displacement means no restoring force.

k \to 0

⟶

F \to 0

An infinitely soft spring exerts no force.

x \to

large⟶ Law breaks downBeyond the elastic limit, permanent deformation occurs and linearity fails.

What if…

What if you stretch beyond the elastic limit?

The spring deforms permanently. F is no longer proportional to x — the material enters plastic deformation.

What if

k \to 0

(infinitely soft)?

The spring offers no resistance. Any displacement produces zero restoring force — like pushing through air.

What if you connect two springs in series?

Effective $k = (k_{1}\cdot k_{2})/(k_{1}+k_{2})$ . Two 100 N/m springs in series act like one 50 N/m spring — softer, not stiffer.

1

Bungee cord spring constant

Given ·

m person:: 70
x:: 4
g:: 9.8

Find · k

Steps

At equilibrium, spring force balances weight: $kx = mg$
Solve for k: $k = mg/x$
$k = 70 \times 9.8 / 4 = 171.5\,\mathrm{N/m}$

Result · At equilibrium:

kx = mg \to k = mg/x = 70 \times 9.8 / 4 = 171.5\,\mathrm{N/m}

2

Car suspension compression

Given ·

k:: 25000
x:: 0.02

Find · F

Steps

Each spring has $k = 25$ ,000 N/m
Road bump compresses spring by $2\,\mathrm{cm} = 0.02\,\mathrm{m}$
$F = kx = 25000 \times 0.02 = 500\,\mathrm{N}$

Result ·

F = kx = 25000 \times 0.02 = 500\,\mathrm{N}

per spring

Simple Harmonic Motion Period→Kinetic Energy→Newton's Second Law→