Hooke's Law & Spring Lab

Explore force, spring constant, displacement, stored elastic energy, and series/parallel springs in real time.

Worked example — Hooke's Law & Spring Lab

Spring constant k	10 N/m
Spring constant k₁	10 N/m
Spring constant k₂	10 N/m
Mass m	0.1 kg
Damping constant c	0.00 N·s/m
Displacement x	-0.45 m

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Every stable well looks like a spring up close 🖖

Hooke's law, F = -kx, describes the linear elastic behavior of materials under deformation. The negative sign represents the restoring force, pushing the system back toward its equilibrium state. At the atomic scale, Hooke's law is a first-order approximation of the Lennard-Jones potential, which governs chemical bonds between atoms. Near the bottom of any stable potential well, the potential energy curve is approximately quadratic, U = 1/2 kx². This quadratic energy state means that any small disturbance will result in simple harmonic motion, where the frequency of oscillation is independent of the amplitude. Hooke's law is thus the mathematical gateway to understanding sound waves, mechanical vibrations, crystal lattices, and quantum harmonic oscillators.

Stiffness, and how springs team up 🖖

Hooke's law just says a spring pushes back in direct proportion to how far you stretch or squeeze it — the spring constant k is simply its stiffness, in newtons per metre. The systems mode reveals a tidy surprise: two identical springs side by side (parallel) double the stiffness, but those same two joined end to end (series) become only half as stiff. Chaining springs makes them softer, not stronger.

Hooke hid his law inside an anagram 🖖

In 1676 Robert Hooke was not ready to reveal his discovery, so he published it as a scrambled Latin anagram: ceiiinosssttuv. Two years later he unscrambled it — ut tensio, sic vis, "as the extension, so the force." He was guarding his priority while racing to build spring-driven watches. That same linear law now sits inside kitchen scales, car suspensions, and the click of a retractable pen.

HOOKE’S LAW — WHICH QUESTION, AND HOW ARE THE SPRINGS ARRANGED?

Which Spring Case Are You In?

A spring carries one number, its stiffness k, and everything else follows from what you ask of it. Pull it and the force is kx. Hold it there and the stored energy is ½kx², which grows with the square of the stretch rather than in step with it. Let go and that same k becomes a clock. And when two springs work together, whether their stiffnesses add or their softnesses do depends entirely on how they are joined.

You want the force — a straight line through the origin F = kx

You want the stored energy — a parabola, so it grows with the square E = ½kx²

You let go — the spring becomes a clock ω₀ = √(k/m)

Two springs side by side — the stiffnesses add k_eq = k₁ + k₂

Two springs end to end — the pair is softer than either 1/k_eq = 1/k₁ + 1/k₂

You want the force — a straight line through the origin

What you know: One spring and one displacement. F = kx: the force is proportional to how far the end has moved from rest, and it pulls back towards rest.

Relation: F = kx

Worked example: k = 100 N/m stretched by x = 0.3 m → F = 100 × 0.3 = 30 N

Open this case: Force vs displacement

You want the force — a straight line through the origin. Force against displacement is a straight line whose slope is the stiffness. One spring and one displacement. F = kx: the force is proportional to how far the end has moved from rest, and it pulls back towards rest. — Force against displacement is a straight line whose slope is the stiffness.

You want the stored energy — a parabola, so it grows with the square

What you know: The work done against a force that grows linearly is the area under that line, which comes to ½kx². Twice the stretch stores four times the energy.

Relation: E = ½kx²

Worked example: k = 150 N/m held at x = −0.4 m → E = ½ × 150 × 0.16 = 12 J; at half that displacement only 3 J

Open this case: Spring energy

You want the stored energy — a parabola, so it grows with the square. Stored energy is the area under the force line: a parabola in the displacement. The work done against a force that grows linearly is the area under that line, which comes to ½kx². Twice the stretch stores four times the energy. — Stored energy is the area under the force line: a parabola in the displacement.

You let go — the spring becomes a clock

What you know: Released from rest, the mass oscillates at ω₀ = √(k/m), whatever distance you pulled it. The damping c only decides how quickly the swing dies away.

Relation: ω₀ = √(k/m)

Worked example: k = 120 N/m with m = 1.5 kg → ω₀ = 8.944 rad/s, so T = 0.702 s and f = 1.42 Hz. With c = 0.3 the damping ratio is ζ = 0.011 — barely damped at all.

Open this case: Damped oscillation

The mass oscillates at a rate set by k and m alone; damping only shrinks the envelope.

Two springs side by side — the stiffnesses add

What you know: Both springs span the same gap, so both stretch by the same x and each contributes its own force. The pair behaves as a single spring of k₁ + k₂.

Relation: k_eq = k₁ + k₂

Worked example: k₁ = 80 N/m and k₂ = 120 N/m in parallel → the pair acts as 200 N/m, so a 0.25 m stretch needs 50 N

Open this case: Parallel springs

Two springs side by side — the stiffnesses add. Both springs stretch by the same amount, so at any x their forces add. Both springs span the same gap, so both stretch by the same x and each contributes its own force. The pair behaves as a single spring of k₁ + k₂. — Both springs stretch by the same amount, so at any x their forces add.

Two springs end to end — the pair is softer than either

What you know: The same force runs through both springs, so each stretches by its own amount and the two extensions add. It is the reciprocals of the stiffnesses that add.

Relation: 1/k_eq = 1/k₁ + 1/k₂

Worked example: Two springs of 150 N/m in series → 150 × 150 / 300 = 75 N/m for the pair. At 0.25 m in total each stretches 0.125 m, and the force is 18.75 N.

Open this case: Series springs

Two springs end to end — the pair is softer than either. The same force passes through both, so the extensions add and the pair is softer. The same force runs through both springs, so each stretches by its own amount and the two extensions add. It is the reciprocals of the stiffnesses that add. — The same force passes through both, so the extensions add and the pair is softer.

References (1)

Insight block 3 — the anagram, and the law it concealed: R. Hooke, Lectures de Potentia Restitutiva, or of Spring. London, 1678 — where "ceiiinosssttuv" is unscrambled to "ut tensio, sic vis".

Problem solved in full

Two 150 N/m springs joined end to end, stretched 0.25 m in total 5 steps
Two 150 N/m springs joined end to end, stretched 0.25 m in total. Find the stiffness of the pair, the force it pulls with and the energy it stores. This is Multiple Springs in the Series Springs configuration, with k₁ = k₂ = 150 N/m and a 0.1 kg mass.
1. Start from what the two springs share. A stretched spring pulls equally at both of its ends, so the tension the first spring hands to the second is the same tension the wall feels. The force is common to both springs; the extension is not.
  
  $F = k_{1} x_{1} = k_{2} x_{2}$
2. Extension is what adds, because the total stretch is the two stretches end to end. Dividing out the common force leaves the reciprocals adding, and reciprocals adding is why the pair comes out softer than either spring on its own.
  
  $x = x_{1} + x_{2} = \frac{F}{k _{1}} + \frac{F}{k _{2}} ⟹ \frac{1}{k _{e q}} = \frac{1}{150} + \frac{1}{150}, k_{e q} = 75 N/m$
3. The pair now behaves as a single 75 N/m spring at the stated 0.25 m. Each spring supplies the full force while doing only half the stretching — the panel rounds the 18.75 N to 18.8 N.
  
  $F = k_{e q} x = 75 \times 0.25 = 18.75 N, x_{1} = x_{2} = \frac{18.75}{150} = 0.125 m$
4. Energy is the area under the force–extension line, so it follows k_eq exactly as the force does. Split between the springs it is two equal shares of 1.172 J. One 150 N/m spring taken to the same 0.25 m would hold 4.688 J, twice as much, because reaching that stretch takes twice the force.
  
  $P E = \frac{1}{2} k_{e q} x^{2} = \frac{1}{2} (75) (0.25)^{2} = 2.344 J = 2 \times \frac{1}{2} (150) (0.125)^{2}$
5. Halving the stiffness does not halve the period. Stiffness sits under a square root, so the pair takes √2 times as long as the 0.162 s the same mass would take on a single spring.
  
  $T = 2 π \frac{m}{k _{e q}} = 2 π \frac{0.1}{75} = 0.229 s = 2 \times 0.162 s$
Answer

The tool prints k_eq = 75.0 N/m, a restoring force of 18.8 N, 2.344 J stored and a period of 0.229 s. Read the series rule backwards and it says something about a single spring: any spring is its own two halves in series, so each half must be twice as stiff as the whole. Stiffness is inversely proportional to free length, and a spring cut in half is not a weaker spring but a stiffer one — which is both the cheap way to firm up a suspension and the usual way to ruin one. It also gives step 2 without algebra: joining two identical springs end to end is nothing more than building one spring of twice the length.

$\frac{1}{k _{e q}} = i \sum \frac{1}{k _{i}} ⟹ k \propto \frac{1}{L}$

Learning path

From one spring to a standing wave

Leads to Resonance

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Physics lessons →