Playground

Interactive spring: drag the mass to stretch/compress and see restoring force update. Spring coils animate.

Variables

SymbolNameSIDimensionRange
FFRestoring Forceoutput
Force exerted by the spring (magnitude)
NMLT^-20 – 500
kkSpring Constant
Stiffness of the spring
N/mMT^-21 – 500
xxDisplacement
Displacement from the equilibrium position
mL0 – 2

Deep dive

Derivation
Empirical law: for small deformations, the restoring force is proportional to displacement. Derived from Taylor expansion of any potential energy minimum: U(x) ≈ U₀ + ½kx².
Experimental verification
Verified with calibrated springs, atomic force microscopes, and crystal lattice vibration measurements.
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
Real-world applications
  • Vehicle suspension systems
  • Seismograph design
  • Atomic bond modeling
  • Mattress and cushion engineering

Worked examples

Bungee cord spring constant

Given:
m_person:
70
x:
4
g:
9.8
Find: k
Solution

At equilibrium: kx = mg → k = mg/x = 70 × 9.8 / 4 = 171.5 N/m

Car suspension compression

Given:
k:
25000
x:
0.02
Find: F
Solution

F = kx = 25000 × 0.02 = 500 N per spring

Scenarios

What if…
  • scenario:
    What if you stretch beyond the elastic limit?
    answer:
    The spring deforms permanently. F is no longer proportional to x — the material enters plastic deformation.
  • scenario:
    What if k → 0 (infinitely soft)?
    answer:
    The spring offers no resistance. Any displacement produces zero restoring force — like pushing through air.
  • scenario:
    What if you connect two springs in series?
    answer:
    Effective k = (k₁·k₂)/(k₁+k₂). Two 100 N/m springs in series act like one 50 N/m spring — softer, not stiffer.
Limiting cases
  • condition:
    x → 0
    result:
    F → 0
    explanation:
    No displacement means no restoring force.
  • condition:
    k → 0
    result:
    F → 0
    explanation:
    An infinitely soft spring exerts no force.
  • condition:
    x → large
    result:
    Law breaks down
    explanation:
    Beyond the elastic limit, permanent deformation occurs and linearity fails.

Context

Robert Hooke · 1676

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

Hook

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.

Dimensions:
lhs:
F → [MLT⁻²]
rhs:
k·x → [MT⁻²]·[L] = [MLT⁻²]
check:
Both sides are [MLT⁻²] = Newton. ✓
Validity: Valid within the elastic limit of the material. Breaks down for large deformations where stress-strain is nonlinear.

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