Spring Constant Calculator

Hooke's law

The restoring force of an ideal linear spring is proportional to displacement from equilibrium. The negative sign indicates that the force opposes the displacement.

F = -k \, x \quad\Leftrightarrow\quad k = (|F|)/(|x|)

Spring constant from a measurement

To determine k, you apply a known force, measure the displacement, and divide. The same k describes the response regardless of the load used.

Simple harmonic motion

Once a mass m is attached, the spring–mass system oscillates with angular frequency ω² = k/m. The full period is

T = (2π)/(\omega) = 2π √(\fracm)k

Energy storage

The elastic potential energy stored in a deformed spring is the integral of F dx from 0 to x:

U = \tfrac12 k x^2

Series and parallel combinations

Springs in parallel add: k_\mathrmeq = k_1 + k_2. Springs in series combine reciprocally: \(k_\mathrmeq = (1/k_1 + 1/k_2)^-1\).

The intuition is worth holding on to: parallel springs share the load, so the assembly is stiffer than either alone; series springs each carry the full load and their extensions add, so the assembly is softer than the softest member. Two identical springs give 2k in parallel and k/2 in series.

Energy Stored in a Spring

Because the force rises linearly with extension, the work done stretching a spring is the area under the force-displacement line — a triangle, giving U = \tfrac12kx^2. Note the square: doubling the extension quadruples the stored energy.

This is the basis of the spring's role as an energy reservoir. A car suspension spring absorbs the kinetic energy of a wheel dropping into a pothole and returns it; the damper is what prevents that energy from simply bouncing back. In a mechanical watch, the mainspring stores the energy for the whole running period, which is why its torque falls as it unwinds and why a fusee or going-barrel is needed to even out the delivery.

U = ∫_0^x kx\,dx = \tfrac12kx^2

Oscillation and the Mass-Spring System

A mass on a spring is the archetypal simple harmonic oscillator. Combining Hooke's law with Newton's second law gives an angular frequency \omega = √(k/m) and hence a period T = 2π√(m/k).

Unlike the pendulum, the period here *does* depend on mass — a heavier mass on the same spring oscillates more slowly. This gives a neat experimental route to k: measure the period for several masses and plot T^2 against m, which should be a straight line of slope 4π^2/k. It also explains why a loaded lorry rides more softly than an empty one: the same suspension stiffness with more mass gives a lower natural frequency.

T = 2π√(\fracm)k, \qquad \omega = √(\frack)m

Limits of Hooke's Law and Practical Notes

Hooke's law is a linear approximation valid only up to the elastic limit. Beyond it the material yields, the spring takes a permanent set, and the measured k from a fresh spring no longer applies. Past the proportional limit but before yield, the force-extension curve also bends away from a straight line, so a single stiffness value becomes an average rather than a constant.

Units are newtons per metre in SI; N/mm (equal to 1000 N/m) is common on engineering data sheets, and lbf/in (about 175.1 N/m) appears in imperial catalogues. When measuring, always work from the spring's free length with no load, take readings both loading and unloading to reveal any hysteresis, and remember that a coil spring quoted with a "preload" already carries force at its installed length — so the displacement in F = kx must be measured from free length, not from the installed position.

F = kx \quad (x \le x_\textelastic limit)

Worked Example

Inputs

• force: 10
• displacement: 0.05
• mass: 0.25

Result: springConstant: 200 period: 0.2221 elasticEnergy: 0.25

Explanation

A 10 N force stretches the spring by 0.05 m, giving k = 10/0.05 = 200 N/m. With a 0.25 kg mass attached, the oscillation period is T = 2π√(0.25/200) ≈ 0.222 s, and the stored elastic energy at 0.05 m is ½·200·(0.05)² = 0.25 J.

Second Scenario

Inputs

• force: 13.5
• displacement: 0.05
• mass: 0.25

Result: springConstant: 200 period: 0.2221 elasticEnergy: 0.25

Explanation

This scenario uses different inputs (force = 13.5, displacement = 0.05, mass = 0.25) to show how changing one variable affects the spring constant result. Run the calculator above with these values to get the exact updated output with step-by-step work.