Physics / Hooke’s law
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EXPERIMENT 12 · PHYSICS LABORATORY

Hooke’s law theory

Understand hooke’s law: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
δ = (1/k)F + b; k = 1/m

Keep the physical quantities, graph variables and units consistent.

01Hooke’s law and linear elasticity

Robert Hooke formulated the principle that for relatively small deformations, the strain or elongation in an elastic material is directly proportional to the applied deforming force. Within this linear regime, doubling the applied force doubles the extension of the spring.

F = kδ
Does Hooke’s law apply to all deformations of a spring?

No. Hooke’s law applies only up to the proportional limit. If the load exceeds the elastic limit, the material undergoes permanent plastic deformation and will not return to its original unstretched length upon unloading.

02The spring constant and stiffness

The spring constant k, also called the stiffness or force constant, measures how much force is required to produce unit extension. It depends on the shear modulus of the wire material, the wire diameter, the coil radius, and the number of active coils.

k = F / δ (units: N m⁻¹ or N/m)
How does cutting a spring in half affect its spring constant?

Cutting a spring in half halves the number of active coils. Since each remaining coil undergoes half the total deformation for a given force, the spring constant of each half doubles (k_half = 2k_full).

03The elastic limit and yield behavior

The proportional limit is the greatest stress a material can sustain without deviating from linear proportionality. Closely following it is the elastic limit, beyond which the spring suffers permanent plastic set and Hookean behavior collapses.

δ_max = F_limit / k
How can an experimenter detect if the elastic limit has been exceeded?

By measuring the pointer position after all weights are removed. If the unloaded pointer position does not return to its initial zero reading, the spring has acquired permanent set.

04Elastic potential energy and work

Because the restoring force increases linearly with displacement from zero to F, the mechanical work done in stretching the spring equals the average force multiplied by extension. This work is stored as elastic potential energy in the twisted wire of the helical coils.

U = ½ k δ² = ½ F δ
Why is the work done ½kδ² rather than kδ²?

The applied force is not constant throughout the deformation; it starts at zero and grows linearly to kδ. Integrating F(x) dx from 0 to δ yields ½kδ², representing the triangular area under the force-extension graph.

05Series and parallel spring combinations

When two identical springs of constant k are linked end-to-end (series), both carry the same load and their extensions add, halving the effective stiffness (k_eff = k/2). When placed side-by-side (parallel), each carries half the load, doubling the stiffness (k_eff = 2k).

1/k_series = 1/k₁ + 1/k₂ · k_parallel = k₁ + k₂
Which arrangement yields greater elongation for a fixed hanging mass?

The series arrangement yields twice the elongation of a single spring (and four times the elongation of a parallel pair) because each spring experiences the full downward weight.

06Static versus dynamic determination of k

The spring constant can be measured statically from load versus extension, or dynamically by timing oscillations of an attached mass M. The dynamic period is T = 2π√((M + m_eff)/k), where m_eff ≈ m_spring / 3 accounts for the distributed inertia of the vibrating coils.

T = 2π √[(M + m_spring/3) / k]
Why does the dynamic period formula include m_spring / 3?

The coils of the spring do not all move with the same amplitude. The top coil is stationary while the bottom coil moves with the suspended mass. Integrating the kinetic energy across all coils reveals that approximately one-third of the spring’s own mass oscillates dynamically.

07Experimental uncertainty and elastic hysteresis

Real physical springs exhibit small mechanical hysteresis where the unloading extension slightly lags behind the loading extension due to internal friction. Parallax errors when reading the scale, pointer zero shifts, and mass manufacturing tolerances also introduce experimental uncertainties.

residual = δ_observed − (m F + b)
How does loading incrementally and unloading decrementally improve experimental rigor?

Taking readings both while adding weights and while removing weights reveals whether hysteresis is present and allows averaging to cancel direction-dependent frictional lags.

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Connect the equations to a measurement.

The ideal teaching model obeys Hooke’s law δ = F/k across the tested loading range. Real laboratory springs often possess initial coil tension, requiring a small threshold force before extension commences, which manifests as a non-zero horizontal intercept. Fitting extension against force yields slope m = 1/k in SI units (m N⁻¹); the spring stiffness constant is recovered from the inverse gradient k = 1/m in N m⁻¹.