# Spring Constant / Hooke's Law Calculator (F = kx)

Calculate spring constant using Hooke's Law F = kx. Find force, displacement, elastic potential energy, and natural frequency. Free physics calculator.

## What this calculates

Hooke's Law states that the force needed to extend or compress a spring is proportional to the displacement: F = kx, where k is the spring constant (stiffness) and x is the displacement from the natural length. This calculator determines the spring constant, restoring force, elastic potential energy (½kx²), and the natural oscillation frequency.

## Inputs

- **Applied Force (F)** (N) — min 0
- **Displacement (x)** (m) — min 0
- **Spring Constant (if known)** (N/m) — min 0

## Outputs

- **Spring Constant (k)** (N/m) — k = F / x
- **Restoring Force** (N) — F = kx (force required or produced)
- **Elastic Potential Energy** (J) — PE = ½kx²
- **Natural Frequency (with 1 kg mass)** (Hz) — f = (1/2π)√(k/m) for m = 1 kg

## Frequently Asked Questions

**Q: What is Hooke's Law?**

A: Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position: F = kx. The constant k (spring constant) measures the spring's stiffness in N/m. A higher k means a stiffer spring that requires more force to stretch. This law holds only within the elastic limit of the material.

**Q: What is elastic potential energy?**

A: Elastic potential energy (PE = ½kx²) is the energy stored in a deformed elastic object like a spring. When you compress or stretch a spring, you do work that is stored as potential energy. When released, this energy converts to kinetic energy. The energy increases with the square of displacement, so doubling the stretch quadruples the stored energy.

**Q: What happens when you exceed the elastic limit?**

A: Beyond the elastic limit, the material deforms permanently and Hooke's Law no longer applies. The spring will not return to its original length when the force is removed. If stretched further, the material reaches its ultimate tensile strength and breaks. Springs are designed to operate well within their elastic limits.

**Q: How is the spring constant related to oscillation frequency?**

A: A mass m on a spring with constant k oscillates at frequency f = (1/2π)√(k/m). Stiffer springs (higher k) produce faster oscillations, and heavier masses (higher m) produce slower oscillations. This is the basis of mechanical clocks, car suspensions, and many vibration systems.

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Source: https://vastcalc.com/calculators/physics/spring-constant
Category: Physics
Last updated: 2026-04-21