Physics I: Mechanics › Oscillations & gravitation › Hooke's law

F = −kxSay it: “the spring force equals minus k x — the spring constant times the stretch, always pulling back toward equilibrium”

Hooke's law

The force law behind every spring — and the seed that grows into all of simple harmonic motion.

Notation on this page: x is measured from the spring's equilibrium position (positive = stretched one way along your chosen axis), and k is the spring constant in N/m.

Where it comes from

Springs fight back when you deform them — and they fight back proportionally. The cleanest way to see it is the experiment Robert Hooke ran in the 1670s: hang known masses from a spring and measure how far it stretches. Here is the data (g = 9.8 m/s²):

Before reading on: a 100 g mass stretches a spring 4.0 cm. You hang 300 g instead. Predict the stretch — double? triple? something else? — then check the table.
100 g → F = 0.98 N
→
x = 0.040 m  ⇒  F/x = 24.5 N/m
Weight = mg = 0.100 × 9.8 = 0.98 N. The ratio F/x is 24.5 N/m.
200 g → F = 1.96 N
→
x = 0.080 m  ⇒  F/x = 24.5 N/m
Double the load, double the stretch — the ratio is unchanged.
300 g → F = 2.94 N
→
x = 0.120 m  ⇒  F/x = 24.5 N/m
Triple the load, triple the stretch. The ratio F/x is a property of the spring, not the load.

That constant ratio is the spring constant k: force per unit stretch. But a ratio alone misses the direction. Stretch the spring to the right (+x) and it pulls back to the left (−x) — the force always opposes the displacement. That restoring character is the minus sign:

F = −kxstretch it +x and it pulls −x; compress it −x and it pushes +x — always home toward equilibriumSay it: “F equals minus k x”

Hooke published it in 1678 as the Latin anagram ceiiinosssttuv, decoded as “ut tensio, sic vis” — as the extension, so the force. Three and a half centuries later it is still the starting point of every oscillation in physics, from guitar strings to car suspensions to molecular bonds.

Derivation

Hooke's law is an experimental law — its “proof” is the measurement itself. We derive the formula from the stretch data above, step by step. The key move is in Step 1: a mass hanging at rest lets us weigh the spring's pull with gravity.

Fspring
=
mg = 0.100 × 9.8 = 0.98 N
Step 1 — weigh the pull. The mass hangs at rest, so the spring's upward pull exactly balances the weight. Gravity becomes our force meter: F = mg.
x
=
0.040 m
Step 2 — measure the stretch. x is measured from the spring's natural length (no load), not from the floor or the ceiling.
k
=
F/x = 0.98 / 0.040 = 24.5 N/m
Step 3 — form the ratio. Repeat: 1.96/0.080 = 24.5, 2.94/0.120 = 24.5. The same k every time — so F = kx in magnitude.
F
=
−kx
Step 4 — fix the direction. Displacement +x (stretched right) → force points −x (pulls left). The minus sign encodes “restoring”. ∎

Where does the law break? Past the spring's elastic limit the data stops being linear — stretch a spring too far and it stays stretched. Hooke's law is the small-deformation law; every real spring has a range where it holds and a point where it doesn't.

How to use it

The procedure, every time:

  1. Find equilibrium. x = 0 is the spring's natural length — mark it before measuring anything.
  2. Choose +x. Stretched toward +x gives F = −kx pointing −x. Write the sign the axis gives you, not the one you feel like.
  3. Solve for the unknown. F = −kx, or k = |F/x|, or x = −F/k. Rearrange before plugging numbers.
  4. Hanging at rest? Forces balance: k|x| = mg (magnitudes). This is how springs become scales — and how you measure an unknown k.
  5. Units check. k is N/m; kx is (N/m)·m = N. If your k comes out in N·m, something is upside down.

Magnitude shortcut

|F| = k|ΔL|when you only need the size of the force — the minus sign just says “toward equilibrium”
Common mistake: setting x to the spring's total length instead of its stretch. A 30 cm spring stretched to 35 cm has x = 5 cm, not 35 cm. Measure from natural length, always.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: k = 200 N/m, stretched 5.0 cm

  1. Inventory. k = 200 N/m, x = +0.050 m (stretched toward +x). Unknown: F.
  2. Write the law. F = −kx.
  3. Substitute. F = −200 × 0.050 = −10 N.
  4. Read the sign. −10 N means 10 N in the −x direction — back toward equilibrium. Units check: (N/m)·m = N ✓
Common mistake: writing F = +10 N, “pushing outward.” The minus sign is the physics: the spring pulls back, never pushes away from equilibrium.
Your turn — k = 150 N/m, stretched 4.0 cm. F = ?

Answer: −6.0 N (6.0 N toward equilibrium). F = −150 × 0.040 = −6.0 N. The negative sign says the force points opposite the stretch — restoring.

Example 2 — weighing k: a 2.0 kg mass hangs at rest, stretching the spring 10 cm

  1. At rest, forces balance. Upward spring pull = downward weight: k|x| = mg.
  2. Weight. mg = 2.0 × 9.8 = 19.6 N. Stretch x = 0.10 m.
  3. Solve. k = 19.6 / 0.10 = 196 N/m.
  4. Sanity check. ~200 N/m is a firm but ordinary spring — plausible for holding 2 kg with a 10 cm sag ✓
Common mistake: dividing by the spring's total hanging length instead of the stretch. x = 0.10 m is how far it stretched past natural length — measure the change, not the length.
Your turn — a 0.50 kg mass hangs at rest, stretching a spring 2.5 cm. k = ?

Answer: 196 N/m. mg = 0.50 × 9.8 = 4.9 N; k = 4.9 / 0.025 = 196 N/m. Same stiffness as Example 2 — k belongs to the spring, not the load.

Example 3 — compression: k = 500 N/m, compressed 3.0 cm

  1. Set the axis. +x to the right. Compression means the end moved left: x = −0.030 m.
  2. Substitute. F = −kx = −500 × (−0.030) = +15 N.
  3. Read the sign. +15 N points +x — the spring pushes the mass back out toward equilibrium ✓
Common mistake: calling compression “positive x” and writing F = −15 N — which claims the spring sucks the mass further inward. Let the axis decide the sign of x; the formula handles the rest.
Your turn — k = 300 N/m, compressed 2.0 cm. F = ?

Answer: +6.0 N. x = −0.020 m, so F = −300 × (−0.020) = +6.0 N — pushing back toward equilibrium, as a compressed spring should.

Before reading on: the same 20 N load hangs from a soft spring (k = 100 N/m) and a stiff spring (k = 400 N/m). Which stretches more — and by what factor?

Example 4 — judgment call: soft vs. stiff spring under a 20 N load

  1. Soft spring. |x| = F/k = 20/100 = 0.20 m.
  2. Stiff spring. |x| = 20/400 = 0.050 m — one quarter the stretch.
  3. The lesson. Stiffer spring (bigger k) → less stretch for the same force. A truck's suspension uses large k so a heavy load barely sags; a soft k would bottom out.
Common mistake: “bigger k means bigger stretch” — reading k as “springiness.” k is stiffness: it resists stretch. Double k, halve x, every time.
Your turn — a 12 N load: spring A (k = 60 N/m) vs. spring B (k = 240 N/m). Stretches?

Answer: A stretches 0.20 m, B stretches 0.050 m. xA = 12/60 = 0.20 m; xB = 12/240 = 0.050 m. B is 4× stiffer, so it stretches 1/4 as far.

Memorization tips

  • Say the minus aloud: “F equals minus k x.” Dropping it in speech is how you drop it on paper.
  • Minus = restoring. The force always points home to x = 0. If your answer's force points away from equilibrium, the sign is wrong — no exceptions.
  • k is stiffness, not springiness. Big k = stiff = hard to stretch. N/m: newtons per metre of stretch.
  • “Ut tensio, sic vis.” As the extension, so the force — Hooke's 1678 one-liner is the law.
  • The doubling test (your 5-second self-check): double the load, double the stretch. If a problem's numbers don't scale that way, you're past the elastic limit — or misreading x.
  • Rebuild from zero: no stretch → no force; a little stretch → a little restoring force. The formula just says that, linearly.

Final challenge

Five mixed questions — computations, units, series/parallel, and the traps, all in one. Score 5/5 and Hooke's law is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the spring constant k?

The stiffness of the spring: the force needed per metre of stretch, in N/m. A larger k means a stiffer spring — more force for the same stretch.

Why is there a minus sign in F = −kx?

The minus sign makes the force restoring: it always points opposite the displacement, back toward equilibrium. Without it the formula would claim the spring pushes itself further outward.

Is x the spring's total length?

No — x is the displacement from the spring's natural (equilibrium) length: how far it is stretched or compressed. A 30 cm spring stretched to 35 cm has x = 5 cm.

Does Hooke's law work for rubber bands?

Only for small stretches. Real materials follow F = −kx up to their elastic limit; stretch a rubber band too far and the force grows faster than linear, then the material deforms permanently.

How do springs in series and parallel combine?

In parallel the k values add (k1 + k2) — each spring stretches the same amount and their forces add. In series 1/k = 1/k1 + 1/k2, so two identical springs in series give k/2.

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