Spring Constant Calculator
How stiff is this spring? Two routes to one k: hang a known weight and divide by the stretch, or time the bounce and let the period name it — and when both run, they check each other.
Spring Constant Calculator
Results recalculate instantly on every keystroke. Nothing you type is transmitted.
What this result does not account for
- Ideal linear spring — no bob-mass correction for the spring's own weight
- Damping ignored; the bounce assumed to keep time
In short: The static route: hang 30.000000 N on the spring and it stretches 0.12 m, so k = F/x = 250.000000 N/m. The dynamic route needs only a stopwatch: a 0.5 kg mass on the same spring bounces with period 0.281 s, and k = 4π²m/T² = 249.986814 N/m — the routes agree within a hundredth of a percent, which is what a healthy spring looks like. That second route is the page's superpower: it needs no known weight, works horizontally, and it is how stiffness is measured when gravity would only get in the way. The same k prices the oscillation: f = (1/2π)√(k/m) = 3.558813 Hz — stiffer springs tick faster, heavier masses slower, both by a square root.
Formula
static: k = F/x · dynamic: k = 4π²m/T² · f = (1/2π)·√(k/m) · T = 2π·√(m/k)
The static route divides force by stretch — Hooke's law read backwards. The dynamic route needs no known weight: a mass bouncing on the spring carries k in its period, since stiffer springs tick faster as √k. The two routes share no inputs, which is exactly why they make such honest witnesses for each other.
Worked Example
- Route one: hang a known weight, read the stretch.
- Route two: time a full bounce of a known mass.
- Run BOTH when you can — the agreement card grades the spring.
- Read the bounce card for the frequency the k promises.
Defaults: static 250.000000 N/m, dynamic 249.986814 N/m — 0.005% apart. The bounce at 0.5 kg: 3.558813 Hz, theory period 0.280993 s (your stopwatch said 0.281).
Strengths & Limits Of This Model
Where this engine is strong
- Two independent routes that cross-check live
- Period route works without any known weight
Where it stops
- No spring-mass correction term
- No damping or resonance modelling
Practical Use Cases
Lab
characterise a spring without a weight set
Engineering
suspension and mount stiffness checks
Teaching
two roads to one constant
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Spring Constant Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
Why two routes at all?
Because they fail differently. The static route needs a known weight and a scale reading; the dynamic route needs a stopwatch and works horizontally, in space, or on any spring too stiff to hang weights from. When both run, their agreement is the cheapest quality check a spring can get.
What does the dynamic formula assume?
An ideal spring — massless, linear — with the bob mass doing all the moving. Real springs add a small effective mass of their own, which is why measured periods run slightly long and k slightly low; for a bob heavy compared with the spring, the error hides below a percent.
Stiffer spring — faster bounce?
Yes, by a square root: f = (1/2π)√(k/m). Quadruple k and the bounce doubles. Quadruple the mass and it halves. That √k is the watchmaker's reason hairsprings are stiff on purpose and the suspension engineer's reason rates are quoted in N/mm.
Why did my static route disagree by more than a percent?
Usually the elastic limit's neighbourhood: beyond the linear region k stops being constant and the static divide silently averages a curve. Also suspect the zero — measure stretch from the rest position with the weight ON, not from the coiled length. The agreement card is exactly the test that catches this.
Can I use weight (kg) directly in the static route?
Convert to newtons first: k = F/x wants force, and a hanging mass supplies F = m·g — a 1 kg hanger is 9.80665 N, not 1. The page keeps the force field honest so the divide is real.
Does the bounce care about gravity?
The period does not: a mass on a spring oscillates at the same frequency sideways on a table or vertically in the lab, because gravity only shifts the rest position, never the restoring gradient. That is why the dynamic route works horizontally and why the two routes can disagree for reasons that are never g.
What is k telling me in plain language?
Newtons per metre of persuasion: 250.000000 N/m means each metre of stretch costs 250 newtons more force. Per-centimetre reading: 2.5 N. It is a slope — the price list of the spring's resistance — which is why parallel combos ADD prices and series ones average down.
How does this pair with the Hooke's law page?
As question to answer: here you MEASURE k from behaviour; there you APPLY k to predict the force at a stretch. Inverse tools sharing one constant and one border — each pointing at the other for the step it does not perform.