Spring Constant & Hooke's Law Calculator

Hooke's Law and Springs Guide

What do I need to know about Hooke's Law?

F = kx. Where F = applied force (N), k = spring constant (N/m), x = extension from natural length (m). Spring constant k represents stiffness — higher k = stiffer spring. Steel coil spring (typical car): k ~30,000-50,000 N/m. Pen spring: k ~100-500 N/m. Bungee cord: k ~50-200 N/m. Elastic limit: beyond a certain extension, the spring no longer returns to its natural length (permanent deformation). Hooke's law only applies within the elastic limit. Most springs obey Hooke's law for extensions up to a certain limit, beyond which the relationship becomes non-linear and the spring may not return to its original length at all.

What should I know about Elastic Potential Energy?

Energy stored in a stretched (or compressed) spring: E = ½kx². The factor of ½ comes from the average force (zero to maximum) acting over the extension. This is the work done stretching the spring. Example: spring k=400 N/m, stretched 0.1m. E = ½ × 400 × 0.01 = 2 J. This energy is released when the spring returns to natural length — useful in: pinball plungers, mousetraps, spring guns, watch escapements, suspension systems. Springs in cars: store the kinetic energy of bumps and release it back gradually, working together with dampers (shock absorbers) to smooth out the ride rather than simply bouncing the energy straight back.

What's the key thing to understand about Simple Harmonic Motion (SHM)?

A mass on a spring oscillates with simple harmonic motion. Period: T = 2π√(m/k). Where m = mass, k = spring constant. Frequency: f = 1/T = (1/2π)√(k/m). Independent of amplitude (assuming Hooke's law holds). Heavier mass: longer period (lower frequency). Stiffer spring (higher k): shorter period. Example: 0.5kg mass on 400 N/m spring. T = 2π√(0.5/400) = 2π × 0.0354 = 0.222s. f = 4.5 Hz. The mass oscillates 4.5 times per second. Angular frequency: ω = 2πf = √(k/m). Equation of motion: a = -(k/m)x, showing the acceleration is always directed back toward the equilibrium position and proportional to displacement — the defining signature of simple harmonic motion.

What should I know about Springs in Series and Parallel?

Combining springs changes effective stiffness. Springs in parallel: stiffness adds. k_total = k₁ + k₂. Example: two 100 N/m springs in parallel: k_total = 200 N/m (stiffer). Springs in series: stiffness combines as: 1/k_total = 1/k₁ + 1/k₂. Example: two 100 N/m springs in series: 1/k_total = 1/100 + 1/100 = 1/50. k_total = 50 N/m (less stiff). Conservation of energy considerations: in vibration isolation (washing machine drum, car suspension), parallel springs provide higher load capacity but less overall travel for the same force, compared to a single spring of the same individual stiffness.