Hooke's Law and Springs

Stretch a rubber band and you can feel it pull back harder the more you extend it. Compress a spring and it pushes back harder as you squeeze it. This instinctive relationship between stretch and force has an elegant mathematical form, first described by Robert Hooke in 1678, and now used in everything from car suspension to atomic force microscopes and gravitational wave detectors. This lesson covers OCR Module 3.4.1 (Springs and elasticity).

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Hooke's Law

For many elastic materials, within a limit, the extension is directly proportional to the applied force:

F = k x

• F is the force applied to the spring (N).
• x is the extension — the change in length from the natural (unstretched) length (m).
• k is the spring constant (also called the stiffness), with SI unit N m⁻¹.

The law was originally stated by Hooke as the Latin anagram "ceiiinosssttuv", which he later published as "Ut tensio, sic vis" — "as the extension, so the force".

Important: Hooke's law assumes the spring is not overstretched. Beyond a certain point (the limit of proportionality), the relationship becomes non-linear.

Exam Tip: Always state Hooke's law in full: "Force is directly proportional to extension, provided the limit of proportionality is not exceeded." Partial statements lose marks.

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The Limit of Proportionality and Elastic Limit

Three related but distinct points on a force-extension graph:

1. Limit of proportionality (P) — the point beyond which F is no longer proportional to x. The graph ceases to be a straight line.
2. Elastic limit (E) — the point beyond which the spring no longer returns to its original length when unloaded (plastic deformation begins).
3. Yield point — where the material begins to deform permanently under a small increase in force (more relevant to metals than springs).

For most springs P and E are very close together and are often used interchangeably at A-Level.

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Force-Extension Graphs

For a spring obeying Hooke's law:

• Plotting F on the y-axis and x on the x-axis, you get a straight line through the origin.
• The gradient is the spring constant k.
• The area under the graph equals the work done stretching the spring (the elastic potential energy stored).

flowchart LR
  A[Apply force F] --> B[Spring extends by x]
  B --> C[Restoring force = kx]
  C --> D[Work done = 0.5 F x]
  D --> E[Stored as elastic PE]

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Worked Example — Finding the Spring Constant

A 2.0 kg mass is hung from a spring. The spring extends by 8.0 cm. Find the spring constant.

• Force F = mg = 2.0 × 9.81 = 19.62 N
• Extension x = 0.080 m
• k = F/x = 19.62 / 0.080 = 245 N m⁻¹

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Elastic Potential Energy

The elastic potential energy stored in a spring when stretched by extension x is equal to the work done against the restoring force. Since the force is not constant (it grows linearly from 0 to F = kx), we cannot simply use W = Fx — we need the average force:

Average force = ½ (0 + kx) = ½ kx Work done = average force × distance = ½ kx × x = ½ k x²

So:

E (elastic) = ½ F x = ½ k x² = F² / (2k)

All three forms are equivalent; choose whichever matches the quantities you know.

Alternatively (and more rigorously), the work done is the area under the force-extension graph — a triangle of base x and height kx, giving area ½ kx². This derivation is worth 2–3 marks on its own in some questions.

Worked Example — Energy in a Stretched Spring

A spring with constant 150 N m⁻¹ is stretched by 0.12 m. Find the stored elastic PE.

E = ½ k x² = ½ × 150 × 0.12² = ½ × 150 × 0.0144 = 1.08 J

If this energy is suddenly released, it can be converted into kinetic energy — for example, propelling a toy dart gun or releasing a catapult.

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Springs in Series

Two springs connected end-to-end are said to be in series. Applying a force F:

• The same force acts through both springs.
• Each extends according to its own spring constant: x₁ = F/k₁, x₂ = F/k₂.
• Total extension: x = x₁ + x₂ = F (1/k₁ + 1/k₂).

So the effective spring constant k_eff of two springs in series satisfies:

1/k_eff = 1/k₁ + 1/k₂

or equivalently k_eff = k₁ k₂ / (k₁ + k₂).

Two identical springs (k each) in series give k_eff = k/2 — softer than either spring alone.

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Springs in Parallel

Two springs sharing the load side-by-side are in parallel:

• Both have the same extension x.
• Each provides its share of the force: F₁ = k₁ x, F₂ = k₂ x.
• Total force: F = F₁ + F₂ = (k₁ + k₂) x.

So:

k_eff = k₁ + k₂