Hooke’s law
For a spring, or any elastic material within its limits, the restoring force is proportional to the displacement from rest:
F = k × x
- k: spring constant or stiffness, in newtons per meter (N/m) or pounds-force per inch (lbf/in)
- x: how far the spring is stretched or compressed
A stiff spring has a large k; a soft spring a small one. 1 lbf/in equals about 175 N/m.
Stored energy
Work done to stretch a spring is stored as elastic potential energy:
E = ½ × k × x²
Doubling the stretch quadruples the stored energy.
Worked example
A spring has a stiffness of 1 lbf/in (175 N/m) and is stretched 2 inches (5.08 cm).
- Force: 1 × 2 = 2 lbf, or 8.90 N.
- Energy: ½ × 175 × 0.0508² = 0.226 J.
- Hanging mass that stretches it that far: a 2 lb weight.
- If that mass bounces on the spring, it oscillates at f = √(g ÷ x) ÷ 2π = 2.21 Hz.
Measuring a spring constant
- Hang the spring and note its length.
- Add a known mass and measure the new length.
- Compute k = m × g ÷ stretch.
- Repeat with several masses and plot force against stretch: the slope of the straight line is k.
Springs in series and in parallel
- In parallel (side by side), stiffnesses add: k = k₁ + k₂.
- In series (end to end), the inverses add: 1/k = 1/k₁ + 1/k₂, so the combination is softer.
Where it applies
Hooke’s law describes car suspensions, mattress coils, scales, trampolines and the tiny deformations of beams and bridges. It fails beyond the elastic limit, where the material deforms permanently, and for materials like rubber, whose force grows non-linearly with stretch.
Natural frequency
A mass m on a spring of stiffness k oscillates at f = (1 ÷ 2π) × √(k ÷ m). With the mass that produces a static stretch x, this simplifies to √(g ÷ x) ÷ 2π, which is why a car with soft suspension bounces slowly.