Hooke's Law and Springs

Spec mapping: OCR H556 Module 3.4 — Materials (Hooke's law $F = kx$ ; force-extension graphs; limit of proportionality and elastic limit; elastic potential energy $\tfrac{1}{2}kx^2$ ; springs in series and parallel). This is the anchor lesson for OCR PAG 2 (Materials practical work). Refer to the official OCR H556 specification document for exact wording.

Stretch a rubber band and it pulls back harder the further you extend it. Compress a spring and it pushes back harder as you squeeze it. This intuitive relationship between deformation and restoring force has an elegant mathematical form, paraphrasing the 1678 work of Robert Hooke (whose anagram-protected discovery decoded to "as the extension, so the force"). It now underlies everything from car suspensions and watch escapements to atomic-force microscopes and the LIGO gravitational-wave detector's test-mass suspensions. This lesson develops Hooke's law, the elastic potential energy stored in stretched springs, and the rules for combining springs in series and parallel.

Key Definition: Hooke's law states that for a spring (or any elastic body) the restoring force is directly proportional to the extension from its natural length, provided the limit of proportionality is not exceeded: $F = k\,x$ , where $k$ is the spring constant (or stiffness). SI unit of $k$ : N m $^{-1}$ .

Hooke's Law

The applied force needed to extend a spring by $x$ from its natural length, in the elastic regime, is:

$F = k\,x$

with:

$F$ in newtons (N).
$x$ in metres (m), measured from the natural (unstretched) length.
$k$ in N m $^{-1}$ — the spring constant (also called stiffness), an intrinsic property of the spring.

The spring exerts an equal-and-opposite restoring force on whatever is stretching it: $F_{\text{restore}} = -kx$ (the negative sign indicates that the restoring force points back toward the natural length). This restoring-force expression underlies simple harmonic motion (SHM, studied later in the course): a mass on a spring oscillates because $F = -kx$ drives an acceleration $a = -kx/m$ proportional and opposite to displacement.

Common Exam Mistake: Stating Hooke's law as just "F equals k x" without the qualifying clause "provided the limit of proportionality is not exceeded". OCR mark schemes routinely award the mark only for the complete statement including the proviso.

A short historical note. Hooke published his discovery in 1678 in De potentia restitutiva with the Latin maxim "Ut tensio, sic vis" — "as the extension, so the force". The principle predates Newton's Principia by nine years and is the founding result of mechanical materials science.

Limit of Proportionality, Elastic Limit and Yield

Three related but distinct landmarks on the force-extension graph:

Limit of proportionality (P): the largest extension for which $F \propto x$ holds. Beyond P the curve ceases to be linear, but the spring may still be elastic.
Elastic limit (E): the largest extension for which the spring still returns to its natural length on unloading. Beyond E, plastic deformation sets in and the spring retains a permanent extension.
Yield point (Y): the extension at which the spring or material begins to deform plastically under a small increase in force. For most metallic springs Y is close to E.

For an ideal Hookean spring P, E and Y are all close together, and they are often treated as synonymous at A-Level. For a more general material (a copper wire, say), they can differ significantly, and the next lesson distinguishes them carefully.

flowchart TD
  A[Apply increasing force to a spring] --> B{Below limit of proportionality}
  B -->|Yes| C[F = kx linear, restoring force exactly proportional]
  B -->|No| D{Below elastic limit}
  D -->|Yes| E[F vs x curved, but still returns to natural length on unload]
  D -->|No| F[Plastic deformation: permanent extension after unload]
  F --> G[Spring constant changes; original behaviour lost]

Force-Extension Graphs

The canonical Hookean force-extension graph plots applied force on the y-axis against extension on the x-axis. In the linear regime:

The graph is a straight line through the origin.
The gradient equals the spring constant $k$ (units: N m $^{-1}$ ).
The area under the graph equals the work done to stretch the spring, which is stored as elastic PE.

For an ideal Hookean spring the loading and unloading curves coincide along the same straight line, and all the work done in stretching is recoverable. For real springs near or beyond P, the unloading curve falls slightly below the loading curve (hysteresis) and a small amount of energy is dissipated as heat each cycle.

Worked Example 1 — Finding the Spring Constant

A $2.0$ kg mass is hung statically from a vertical spring. The spring extends by $8.0$ cm beyond its natural length. Find the spring constant.

At equilibrium, the spring's restoring force balances gravity:

$F = m g = 2.0 \times 9.81 = 19.6\;\text{N}$

$k = \frac{F}{x} = \frac{19.6}{0.080} = 245\;\text{N m}^{-1}$

Three sanity-check observations: (i) the units work out as N m $^{-1}$ , which is the SI unit of stiffness; (ii) $k \sim 200$ N m $^{-1}$ is typical of a steel coil spring of a few cm long and 1-mm wire diameter; (iii) the answer is independent of the gravitational field strength only if the same calculation is performed at a place with different $g$ — at the Moon's $g \approx 1.62$ N kg $^{-1}$ , the same $2$ kg mass would extend the spring by only $1.32$ cm.

Elastic Potential Energy

The elastic potential energy stored in a spring when it is stretched by extension $x$ equals the work done against the restoring force. Crucially, the force is not constant — it grows linearly from $0$ at zero extension to $F = kx$ at full extension. We therefore use the average force over the stretch:

$F_{\text{avg}} = \tfrac{1}{2}(0 + kx) = \tfrac{1}{2}kx$

Work done by the applied force (and stored as elastic PE):

$E_{\text{elastic}} = F_{\text{avg}} \times x = \tfrac{1}{2}kx \times x = \tfrac{1}{2}kx^2$

So three equivalent forms:

$E_{\text{elastic}} = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^2 = \frac{F^2}{2k}$

Choose whichever matches the quantities you know.

A more rigorous derivation, using calculus: the work done in stretching from $0$ to $x$ against a restoring force $F(x') = -kx'$ is:

$W = \int_0^x F(x')\,dx' = \int_0^x kx'\,dx' = \tfrac{1}{2}kx^2$

The integral $\int kx'\,dx' = \tfrac{1}{2}kx'^2$ is precisely the area under the linear force-extension graph — a triangle of base $x$ and height $kx$ , so area $\tfrac{1}{2}x \times kx = \tfrac{1}{2}kx^2$ . The geometric and calculus pictures agree.

Common Exam Mistake: Writing $E = kx^2$ without the factor of $\tfrac{1}{2}$ , or writing $E = \tfrac{1}{2}Fx$ when the force is given as a peak rather than an average and then mis-adapting elsewhere. The factor $\tfrac{1}{2}$ comes from the average force, since $F$ varies linearly from $0$ to $kx$ .

Worked Example 2 — Energy in a Stretched Spring

A spring of constant $k = 150$ N m $^{-1}$ is stretched by $x = 0.12$ m. Find the stored elastic PE.

$E = \tfrac{1}{2}kx^2 = \tfrac{1}{2} \times 150 \times 0.12^2 = \tfrac{1}{2} \times 150 \times 0.0144 = 1.08\;\text{J}$

If the spring is suddenly released against a mass, this $1.08$ J becomes kinetic energy. For a small mass of $0.050$ kg, $v = \sqrt{2 \times 1.08 / 0.050} = 6.6$ m s $^{-1}$ — the principle behind spring-powered toys, mousetrap cars and pinball plunger launchers.

Springs in Series

Two springs connected end-to-end form a series combination. Applying a force $F$ to the bottom of the lower spring:

The same tension $F$ acts through both springs (Newton III at the join).
Each extends according to its own stiffness: $x_1 = F/k_1$ , $x_2 = F/k_2$ .
Total extension: $x = x_1 + x_2 = F\left(\frac{1}{k_1} + \frac{1}{k_2}\right)$ .

The effective spring constant $k_{\text{eff}}$ of the series combination satisfies:

$\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2}\quad\text{equivalently}\quad k_{\text{eff}} = \frac{k_1 k_2}{k_1 + k_2}$

Two identical springs (each of stiffness $k$ ) in series give $k_{\text{eff}} = k/2$ — softer than either spring alone.

A useful mental model: each spring carries the full tension, and contributes its own extension. The two extensions add, but the force does not.

Springs in Parallel

Two springs sharing the load side-by-side form a parallel combination. Applying a force $F$ across the parallel pair:

Both springs extend by the same amount $x$ (constrained by the shared end-plate).
Each contributes its share of the load: $F_1 = k_1 x$ , $F_2 = k_2 x$ .
Total force: $F = F_1 + F_2 = (k_1 + k_2)\,x$ .

So:

$k_{\text{eff}} = k_1 + k_2$

Two identical springs in parallel give $k_{\text{eff}} = 2k$ — stiffer than either spring alone. This is why car-suspension struts use multiple parallel springs to handle large vehicle weights; why insect leg-joints stack several elastic cuticular elements; why drum suspension brushes use clustered parallel bristles to maintain contact.

Common Exam Mistake: Adding spring constants in series ( $k_{\text{eff}} = k_1 + k_2$ ) or combining them reciprocally in parallel. The convention is exactly opposite to resistor combinations in electrical circuits: resistors add in series, springs add in parallel. A useful memory aid: "Springs in parallel act like resistors in series — both ways, the total quantity is the sum of the components".

Worked Example 3 — Combined Springs

Two springs have constants $k_1 = 200$ N m $^{-1}$ and $k_2 = 300$ N m $^{-1}$ . Find the effective stiffness in (a) series and (b) parallel.

(a) Series: $\frac{1}{k_{\text{eff}}} = \frac{1}{200} + \frac{1}{300} = \frac{3 + 2}{600} = \frac{5}{600}$ , so $k_{\text{eff}} = 120$ N m $^{-1}$ .

(b) Parallel: $k_{\text{eff}} = 200 + 300 = 500$ N m $^{-1}$ .

The parallel combination is more than four times stiffer than the series combination. The series combination is even softer than the softer of the two springs alone — this is non-intuitive at first sight (adding any spring should make the system stiffer, no?) but the geometric series identity shows that adding a softer spring in series can only further increase the total extension for a given force.

Worked Example 4 — Energy in Parallel Springs

Two identical springs ( $k = 80$ N m $^{-1}$ ) are arranged in parallel and stretched by $x = 0.050$ m. Find the total elastic PE stored.

Method 1 — combined effective stiffness:

$k_{\text{eff}} = 80 + 80 = 160$ N m $^{-1}$ , so $E = \tfrac{1}{2} k_{\text{eff}} x^2 = \tfrac{1}{2} \times 160 \times 0.0025 = 0.20$ J.

Method 2 — sum over individual springs:

Each spring stores $\tfrac{1}{2} k x^2 = \tfrac{1}{2} \times 80 \times 0.0025 = 0.10$ J. Total: $2 \times 0.10 = 0.20$ J. ✓

Both methods give the same answer, as they must — energy is additive.

Worked Example 5 — Spring-Launched Mass

A $0.10$ kg mass is held against a horizontal spring of stiffness $250$ N m $^{-1}$ , compressing the spring by $0.080$ m. It is released on a smooth surface. Find the speed of the mass at the instant the spring returns to its natural length.

Conservation of energy: elastic PE → KE.

$\tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2 \Rightarrow v^2 = \frac{kx^2}{m} = \frac{250 \times 0.0064}{0.10} = 16$

$v = 4.0\;\text{m s}^{-1}$

This is the principle behind spring-loaded launchers — pinball plungers, mousetrap cars, ballistic-pendulum demonstration apparatus. The same energy balance underpins SHM: at the equilibrium point of a mass-spring oscillator, all the energy is kinetic; at the extrema, all elastic.

Measuring the Spring Constant — OCR PAG 2

A standard part of OCR PAG 2 (Materials) is determining a spring constant by loading and measuring. The procedure:

Set-up: Suspend the spring vertically from a fixed clamp; attach a hanger to the lower end; place a metre rule alongside with a fiducial pointer at the lowest coil.
Datum: Record the natural-length position $\ell_0$ with the spring under only the hanger's own weight (or zero load if the hanger is massless).
Load: Add masses in increments (typically $50$ g or $100$ g steps); after each addition, allow oscillations to die out and record the new length $\ell$ . Calculate extension $x = \ell - \ell_0$ .
Unload: Reverse the procedure, removing masses one at a time and recording extension. Loading and unloading curves should coincide if the spring is elastic; any divergence indicates plastic deformation or hysteresis.
Plot $F = mg$ on the $y$ -axis against $x$ on the $x$ -axis. Gradient = $k$ .
Repeat for at least two trials and report mean and uncertainty.

Good practice:

Measure extension directly, not total length — this eliminates the systematic error from misreading the natural length.
Keep loads well below the limit of proportionality (typically $< 50\%$ of the manufacturer's spec); if the graph stops being linear, stop adding masses.
Use a metre rule with a fiducial marker at the bottom of the spring; read to the nearest millimetre, with the eye level with the marker to avoid parallax error.
Repeat each reading multiple times to reduce random error.
Include uncertainties in $F$ (from mass tolerance, typically $\pm 0.5\%$ ) and in $x$ (typically $\pm 0.5$ mm for a ruler, less for a Vernier).

Non-Hookean Materials

Not all elastic materials obey Hooke's law. Rubber is the canonical counter-example:

At small extensions (say, $< 100\%$ of natural length), rubber is approximately linear.
At large extensions, the curve becomes non-linear and then suddenly very steep as the polymer chains become fully uncoiled — the so-called strain-stiffening region.
On unloading, rubber follows a different curve below the loading curve. The enclosed area is the energy dissipated as heat in the rubber — hysteresis.

flowchart LR
  A[Hookean spring: linear F vs x] --> B[Loading and unloading curves coincide]
  B --> C[All elastic PE recoverable on unload]
  D[Rubber band: non-linear F vs x] --> E[Loading curve above unloading curve]
  E --> F[Hysteresis loop: area = heat dissipated per cycle]
  F --> G[Rubber warms on cycling]

Hooke's Law and Springs

Hooke's Law and Springs

Hooke's Law

Limit of Proportionality, Elastic Limit and Yield

Force-Extension Graphs

Worked Example 1 — Finding the Spring Constant

Elastic Potential Energy

Worked Example 2 — Energy in a Stretched Spring

Springs in Series

Springs in Parallel

Worked Example 3 — Combined Springs

Worked Example 4 — Energy in Parallel Springs

Worked Example 5 — Spring-Launched Mass

Measuring the Spring Constant — OCR PAG 2

Non-Hookean Materials

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