Hooke's Law states that the restoring force exerted by a spring is proportional to its extension or compression from its natural length: F = kx, where F is the force in newtons, k is the spring constant in N/m, and x is the displacement in metres. The law holds only within the elastic limit — beyond that the material deforms permanently.

How do I find the spring constant k?

Rearrange F = kx to get k = F ÷ x. Hang a known mass m from the spring, measure the extension x, and apply F = mg (g = 9.81 m/s²). For example, a 0.5 kg mass stretching a spring by 2.5 cm gives k = (0.5 × 9.81) / 0.025 = 196.2 N/m.

What is elastic potential energy and how is it calculated?

When a spring is stretched or compressed by x metres it stores elastic potential energy E = ½kx². Doubling the displacement quadruples the stored energy because of the x² term. The energy is released as kinetic energy when the spring returns to its natural length — this is the engine of simple harmonic motion.

How are springs in series different from springs in parallel?

Springs in series share the same force but each deflects independently, giving a softer combined spring: 1/k_eq = 1/k₁ + 1/k₂ + … Springs in parallel share the same displacement but each carries part of the load, giving a stiffer result: k_eq = k₁ + k₂ + … Two identical 1000 N/m springs in series give 500 N/m; in parallel they give 2000 N/m.

What is the period of a spring-mass oscillation?

For a mass m on a spring with constant k undergoing frictionless simple harmonic motion the period is T = 2π√(m/k). Doubling the mass increases T by √2; quadrupling the stiffness halves T. Crucially the period does not depend on amplitude — a small and a large oscillation take exactly the same time per cycle (within the elastic limit).

Does Hooke's Law apply to all springs?

Only up to the elastic limit (or proportionality limit). Below that limit stress is proportional to strain and the spring returns exactly to its original shape. Beyond it the spring yields permanently. The calculator models the linear region; if you are working with large deflections or plastic deformation you need a non-linear constitutive model instead.

Hooke's Law Calculator

Hooke’s Law is one of the most useful relationships in classical mechanics and materials science. It governs everything from the suspension spring in a car to the atomic bonds in a crystal lattice, and its elegant simplicity — F = kx — hides a surprising amount of practical depth. This calculator exposes four distinct modes so you can solve whichever aspect of spring behaviour you need, with unit conversion, step-by-step working and copy-to-clipboard output throughout.

What the calculator covers

Basic F = kx lets you solve for any one of the three variables given the other two. Enter any combination of force, spring constant and displacement in your preferred unit (N, kN, lbf, kgf; m, cm, mm, in, ft; N/m, N/mm, kN/m, lbf/in, lbf/ft) and the result is displayed instantly with a complete algebraic trace.

Series and parallel combinations handle up to five springs at once. Series springs share the same tension but each adds its own deflection — the equivalent stiffness is always softer than the weakest individual spring. Parallel springs share the same displacement — each carries a fraction of the load and the equivalent stiffness is the sum of all constants.

Elastic potential energy (E = ½kx²) quantifies the work done compressing or extending a spring. The quadratic dependence on x means a small extra extension stores disproportionately more energy — doubling x stores four times as much.

Oscillation (SHM) treats the ideal spring-mass system undergoing simple harmonic motion. Given k, mass and amplitude the tool returns angular frequency (ω = √(k/m)), frequency (f = ω/2π), period (T = 2π√(m/k)), peak velocity (v_max = ωA), peak acceleration (a_max = ω²A) and total mechanical energy (E = ½kA²).

How it works

All arithmetic runs in JavaScript inside your browser — nothing is sent to any server. Inputs are converted to SI units internally (N, m, kg, J) before any calculation, then converted back to your chosen display unit for the result. This avoids cascading rounding errors and means you can freely mix unit systems between inputs.

The “Show working” trace expands to reveal each algebraic substitution in sequence, making it straightforward to copy the method into a lab report or homework solution.

Worked example

A car suspension spring has a stiffness of 25 kN/m. The car body rests on four springs and the vehicle mass is 1 400 kg.

Load per spring: F = (1400 × 9.81) / 4 = 3 433.5 N
Static compression: x = F/k = 3 433.5 / 25 000 = 0.137 m (13.7 cm)
Elastic PE per spring: E = ½ × 25 000 × 0.137² = 234.7 J
Natural frequency: f = (1/2π)√(25 000 / 350) ≈ 1.35 Hz (350 kg per corner)

A frequency near 1–1.5 Hz matches the comfortable ride-frequency target for passenger cars. Enter these numbers in the calculator’s Oscillation tab to verify each figure and explore how a stiffer or softer spring shifts the frequency.

Formula reference

Quantity	Formula	SI units
Restoring force	F = kx	N
Spring constant	k = F/x	N/m
Displacement	x = F/k	m
Elastic PE	E = ½kx²	J
Series equivalent	1/k_eq = Σ(1/kᵢ)	N/m
Parallel equivalent	k_eq = Σkᵢ	N/m
Angular frequency	ω = √(k/m)	rad/s
Period	T = 2π√(m/k)	s
Frequency	f = (1/2π)√(k/m)	Hz
Peak velocity	v_max = ωA	m/s
Peak acceleration	a_max = ω²A	m/s²
Total SHM energy	E = ½kA²	J

All calculations are 100% client-side. No figures you enter are uploaded or stored.