Hooke’s law for spring force

Hang small loads from a spring in succession and measure extension relative to its unloaded length. If twice the tensile load gives twice the extension, this suggests a linear range. Stiffness describes the relation between these quantities, rather than a universal property of every spring under arbitrary loading. Signs enter only after choosing a positive direction.[1]

The equation above uses F as positive force magnitude and x as the magnitude of the length change. The spring constant or stiffness k is also sometimes denoted D. A spring with $k=80\,\mathrm{N\,m^{-1}}$ needs a holding force of 0.8 N for a 1 cm change and 1.6 N for 2 cm in the linear model. A larger k means more force at the same displacement.

For a signed length change x relative to the natural unloaded length, spring force is $F_x=-kx$, opposing extension or compression. For a horizontal spring without other constant forces, natural length also defines equilibrium. A vertically loaded spring differs: its force is nonzero at equilibrium and balances weight. Only net force becomes $F_{\mathrm{net}}=-ky$ when y is measured from loaded equilibrium.[1]

The unit N/m describes the increase in force per metre of additional extension within the linear range. Absolute spring length l is not displacement. First measure unloaded length l₀ and calculate $x=l-l_0$. A preloaded spring may carry force F₀ at a chosen operating point. The local linear relation is then $F_{\mathrm{hold}}-F_0=k(x-x_0)$. Forcing a line through the origin would miss that preload.

Original measurement check: 2 N at 8 mm and 5 N at 20 mm give $k=\frac{\Delta F_{\mathrm{hold}}}{\Delta x}=\frac{3\,\mathrm N}{0.012\,\mathrm m}=250\,\mathrm{N\,m^{-1}}$. The second point checks this with $kx=250\cdot0.020\,\mathrm N=5\,\mathrm N$. Multiple measurements show whether constant slope is appropriate; loading and unloading curves can reveal further deviations.

Convert stiffness in N/mm before calculating with metres: $1\,\mathrm{N/mm}=1000\,\mathrm{N/m}$. The measured slope of 250 N/m used here equals 0.25 N/mm. An extension of 20 mm gives the same holding-force magnitude of 5 N with either unit; the spring force points in the opposite direction.

Hooke’s law is a material and component model; Newton’s equation of motion alone does not imply it. When measurements show constant slope over a range, $\Delta F_{\mathrm{hold}}=k\Delta x$ holds there. With no preload and the unloaded state as zero, $F_{\mathrm{hold}}=kx$ follows. This reasoning uses observed linearity and does not prove every spring is linear under every load.

The second example considers the loaded vertical equilibrium position. For a mass–spring oscillator, subsequent motion is measured relative to that position rather than the unloaded spring length.[1]

The elastic energy stored in a spring follows from the work done in extending or compressing it. During slow stretching, holding force rises from zero to kx. Work is therefore the triangular area in a force–displacement plot, not the rectangle of final force times displacement. Stored energy is positive for extension and compression when the relaxed state is chosen as zero energy.[2]

For the spring in the first example, average holding force is 5.6 N. Over 0.035 m this gives $5.6\cdot0.035\,\mathrm J=0.196\,\mathrm J$. Final force times distance gives 0.392 J, twice the correct value. Stretching a spring that is already extended changes energy by $\frac12k(x_2^2-x_1^2)$, not $\frac12k(x_2-x_1)^2$.

The following relations follow directly from the linear model for ideal springs and massless connections. Parallel springs undergo the same extension, so their forces add. In static equilibrium, series springs carry the same force magnitude, so their extensions add.

Original comparison: 120 N/m and 180 N/m give 300 N/m in parallel and 72 N/m in series. Under 6 N, the parallel arrangement extends by 2 cm. In series, extensions are 5 cm and about 3.33 cm, totalling about 8.33 cm. The different force and displacement sharing explains the result.

Elastic means deformation recovers after unloading; linear means constant slope. An elastic but curved force–displacement relation is therefore not necessarily described by constant k. Do not simply extrapolate the line through permanent deformation, coil contact or other geometric limits.[3]

With $F_x=-kx$ and no additional forces, Newton’s law gives $m\ddot x=-kx$. This describes undamped harmonic motion with $\omega=\sqrt{\frac km}$. Real dampers add forces often dependent on velocity. A spring alone is therefore not a complete shock-absorber model. Vertical gravity shifts equilibrium without changing frequency in this ideal model.[1]

First choose whether the coordinate is measured from natural length or loaded equilibrium. Decide whether F means a magnitude, signed spring force or net force. In a graph of external holding force against length change, slope in the linear range is k; signed spring force has slope −k. Reversing the axes gives slopes $\frac1k$ and $-\frac1k$, respectively. Also check units, prestretch and the measured linear range.

The original force–displacement diagram above shows holding force, restoring force and stored energy. A numerical calculation requires stated k or measured data; a schematic drawing does not supply calibrated stiffness. The FAQ below covers common questions about signs and limitations.

Linear spring force is proportional to displacement from unloaded equilibrium. Restoring force opposes displacement, while quasistatic holding force follows it. First check the elastic linear range; total spring length is not displacement.