Hooke's law

Squeeze a rubber ball, tug a bedspring, or lean on a steel ruler until it bows — in every case, the harder you push, the more it deforms, and the moment you let go it springs back. Robert Hooke noticed this in 1676 and hid his discovery in a Latin anagram (ceiiinosssttuv) before revealing its solution years later: ut tensio, sic vis — "as the extension, so the force." Strip away the theatrics and Hooke's law says something almost embarrassingly simple: stress is directly proportional to strain, as long as you do not push too hard. That one proportionality is the hinge on which the entire chapter on Mechanical Properties of Solids turns. Once you know stress equals a constant times strain, every "modulus" you will meet next chapter — Young's, bulk, shear — turns out to be nothing more than that same constant, measured for a different kind of squeeze or stretch. Engineers use it to decide how thick a bridge girder must be before it sags dangerously; instrument makers use it to calibrate the coiled spring in a weighing machine; and your own school's swinging door survives a thousand slams a day because a metal hinge somewhere is obeying exactly this law, elastically, every single time. This section builds Hooke's law from the stress-strain relation you have already met, walks through the one derivation that matters (Young's modulus from a stretched wire), reworks the two worked numbers you saw in Pass 1 with a competitive-exam flavour, and pins down exactly where students lose marks: confusing the elastic limit with the breaking point, forgetting that the modulus carries the units of stress, and assuming one number describes a material for every kind of deformation.

Key takeaways

Hooke's law: within the elastic limit, stress is directly proportional to strain, $\sigma \propto \varepsilon$ , so $\sigma = k\varepsilon$ for some constant $k$ .
The proportionality constant $k$ is the modulus of elasticity; for longitudinal stretching it is called Young's modulus $Y = \dfrac{FL}{A\,\Delta L}$ .
Since strain is a dimensionless ratio, the modulus of elasticity has the same units as stress — pascal (Pa) — never units of force or length.
Hooke's law holds only on the initial straight-line part of the stress-strain graph; beyond the elastic limit / proportional limit, stress and strain stop being proportional even though the material may still be elastic.
For a spring under load, the same law is written $F = k_s x$ , with $k_s$ the spring's force constant — stiffer springs need more force per unit extension.
Elastic potential energy stored per unit volume, $u = \tfrac12\sigma\varepsilon = \tfrac12 Y\varepsilon^2$ , is a direct consequence of the linear stress-strain relation and is the area under the straight-line graph.
Key idea: within the elastic limit, stress and strain rise together in fixed proportion, $\sigma = Y\varepsilon$ , and the constant of proportionality is the material's modulus of elasticity — the single number that predicts how much any solid will stretch, compress, or shear under a known load.

The core insight

Within the elastic limit, stress is directly proportional to strain ( $\sigma = Y\varepsilon$ for stretching), and the constant of proportionality is the material's modulus of elasticity, measured in the same units as stress.

Where it fits in the syllabus

Hooke's law is prescribed in the CBSE Class 11 unit Mechanical Properties of Solids, sitting directly between stress-and-strain (which defines the two quantities) and the stress-strain curve (which shows graphically where the law holds and where it breaks down). The board syllabus asks for the statement $\sigma \propto \varepsilon$ , the identification of the modulus of elasticity as the constant of proportionality, and the specific case of Young's modulus $Y = FL/(A\,\Delta L)$ with full numerical problem-solving. Shear (rigidity) modulus is prescribed only as a qualitative idea — no numerical shear problems are expected — and bulk modulus is covered fully as a separate node. Anything about the elastic tensor, anisotropic materials, or non-linear elasticity theory belongs to university coursework and is deliberately kept out here.

The physical picture

Picture a wire of natural length $L$ and cross-sectional area $A$ , hung vertically and loaded with a weight $F$ at its lower end. The wire stretches by a small amount $\Delta L$ . Two ratios summarise what is happening: the stress, which measures how concentrated the pulling force is on the material's cross-section, and the strain, which measures how much the wire has deformed relative to its original size.

\sigma = \frac{F}{A}, \qquad \varepsilon = \frac{\Delta L}{L}

where: $\sigma$ is the longitudinal stress (Pa), $F$ is the applied force (N), $A$ is the cross-sectional area (m squared), $\varepsilon$ is the strain (dimensionless), $\Delta L$ is the extension (m), and $L$ is the original length (m). Hooke's law is the experimental claim that, for small deformations, these two ratios are simply proportional to each other:

\sigma = Y\,\varepsilon \qquad\Longrightarrow\qquad Y = \frac{\sigma}{\varepsilon} = \frac{FL}{A\,\Delta L}

where: $Y$ is Young's modulus, the constant of proportionality for longitudinal stretching. Because $\varepsilon$ is a pure ratio with no units, $Y$ must carry the same units as $\sigma$ — pascal (Pa) or newton per square metre — never the units of force alone or length alone. A physical reading: a large $Y$ means a material is "stiff" and needs an enormous stress to produce even a tiny strain (steel, $Y \approx 2.0\times10^{11}\ \text{Pa}$ ); a small $Y$ means the material is "soft" and stretches noticeably under modest stress (rubber, $Y$ around $10^6$ to $10^7\ \text{Pa}$ ). The same idea, written for a coiled spring rather than a straight wire, becomes the familiar restoring-force law $F = -k_s x$ , where $k_s$ is the spring's force constant and the minus sign records that the spring always pulls or pushes back toward its natural length. The magnitude form $F = k_s x$ is what you use for numerical problems.

Figure

A stress strain graph showing a straight line through the origin up to the elastic limit point followed by a curved region up to the yield point and fracture point, with the straight portion labelled as the region where Hookes law holds

The graph makes the boundary explicit: the straight portion from the origin up to the elastic limit is where Hooke's law is valid, and its slope is numerically equal to the modulus of elasticity. Past the elastic limit the curve bends — the material may still return to its original shape if unloaded (still "elastic" in the everyday sense) but stress and strain are no longer proportional, so Hooke's law itself has stopped applying even though elasticity has not.

Derivation AYoung's modulus from a loaded wire

The goal here is to connect four measurable, everyday quantities — the force hung on a wire, its cross-sectional area, its original length, and the extension it produces — into a single material constant that predicts the wire's behaviour under any similar load, using nothing beyond the definitions of stress, strain, and Hooke's law's proportionality. This is the one derivation the CBSE syllabus expects you to reproduce from scratch, because every later modulus (bulk, shear) is built the same way.

Step 1. Start from the two definitions already established for a stretched wire: the longitudinal stress is the force per unit cross-sectional area, $\sigma = F/A$ , and the longitudinal strain is the fractional change in length, $\varepsilon = \Delta L / L$ . Both are measurable in a simple laboratory setup — hang a known weight, measure the wire's diameter with a screw gauge, and measure the extension with a travelling microscope or vernier arrangement.

Step 2. Apply Hooke's law as the experimental postulate: within the elastic limit, $\sigma$ is directly proportional to $\varepsilon$ , so we may write $\sigma = k\,\varepsilon$ for some constant $k$ that depends only on the material (and, for shear or volume strain, on the type of deformation) and not on the specific values of $F$ , $A$ , $L$ , or $\Delta L$ individually.

Step 3. For the specific case of longitudinal stretching, this constant $k$ is named Young's modulus $Y$ , giving $\sigma = Y\varepsilon$ . Substituting the definitions from Step 1 gives $\dfrac{F}{A} = Y\,\dfrac{\Delta L}{L}$ .

Step 4. Rearrange for $Y$ : $Y = \dfrac{FL}{A\,\Delta L}$ . This is the working formula for every Young's-modulus problem — plug in the load, the original length, the cross-sectional area, and the measured extension, and out comes a single number characteristic of the material, independent of how long or thick the particular wire sample happened to be. Consistency check: because $\varepsilon$ is dimensionless, $Y$ carries the same SI unit as $\sigma$ , the pascal — confirmed by the formula, since $F/A$ already has units of pressure and dividing by the dimensionless ratio $\Delta L/L$ changes nothing.

Worked examples

Board level - Young's modulus of a copper wire

A copper wire of length $1.8\ \text{m}$ and area of cross-section $0.8\ \text{mm}^2$ obeys Hooke's law. A load of $F = 40\ \text{N}$ produces an extension of $0.5\ \text{mm}$ . Find (a) the stress, (b) the strain, and (c) the modulus of elasticity (Young's modulus) for copper from this data.

Step 1. Convert every quantity to SI units first: $A = 0.8\ \text{mm}^2 = 0.8\times10^{-6}\ \text{m}^2$ , $\Delta L = 0.5\ \text{mm} = 0.5\times10^{-3}\ \text{m}$ , and $L = 1.8\ \text{m}$ needs no conversion.

Step 2. Compute the stress using $\sigma = F/A$ :

\sigma = \frac{40}{0.8\times10^{-6}} = 5.0\times10^{7}\ \text{Pa}

Step 3. Compute the strain using $\varepsilon = \Delta L / L$ :

\varepsilon = \frac{0.5\times10^{-3}}{1.8} = 2.78\times10^{-4}

Step 4. Divide stress by strain to get the modulus of elasticity, $Y = \sigma/\varepsilon$ :

Y = \frac{5.0\times10^{7}}{2.78\times10^{-4}} \approx 1.80\times10^{11}\ \text{Pa}

The Young's modulus of this copper sample is approximately $1.80\times10^{11}\ \text{Pa}$ , which matches the accepted textbook value for copper closely enough to confirm the wire was indeed behaving elastically, well within its proportional limit, when this load was applied.

Competitive level - comparing two springs by force constant

Two springs, A and B, are tested by hanging loads and measuring extension. Spring A extends $2\ \text{cm}$ under a $10\ \text{N}$ load; spring B extends $5\ \text{cm}$ under the same $10\ \text{N}$ load. Both obey Hooke's law in this range. (a) Which spring is stiffer, and what is the ratio of their force constants? (b) If both springs are now combined in series and the same $10\ \text{N}$ load is hung from the combination, find the total extension.

Step 1. Recall the spring form of Hooke's law, $F = k_s x$ , so each force constant is $k_s = F/x$ . Convert extensions to metres first: $x_A = 0.02\ \text{m}$ , $x_B = 0.05\ \text{m}$ .

Step 2. Compute each force constant:

k_A = \frac{10}{0.02} = 500\ \text{N/m}, \qquad k_B = \frac{10}{0.05} = 200\ \text{N/m}

Step 3. Compare: $k_A / k_B = 500/200 = 2.5$ . Spring A needs 2.5 times the force per unit extension that spring B needs, so spring A is the stiffer spring.

Step 4. For part (b), recall that springs in series share the same force but add extensions, which is equivalent to adding their compliances: $\dfrac{1}{k_{\text{series}}} = \dfrac{1}{k_A} + \dfrac{1}{k_B} = \dfrac{1}{500} + \dfrac{1}{200} = \dfrac{2}{1000} + \dfrac{5}{1000} = \dfrac{7}{1000}$ , giving $k_{\text{series}} = \dfrac{1000}{7} \approx 142.9\ \text{N/m}$ .

Step 5. The total extension under the same $10\ \text{N}$ load is $x = F/k_{\text{series}} = 10 / 142.9 \approx 0.070\ \text{m} = 7\ \text{cm}$ , which is simply the sum $x_A + x_B = 2\ \text{cm} + 5\ \text{cm} = 7\ \text{cm}$ , exactly as expected for a series combination.

The force-constant ratio is 2.5, and the series combination stretches by 7 cm under the same 10 N load. Timing note: this two-part question is designed to be solved in under 3 minutes in a competitive exam — recognise the series-spring shortcut ( $x_{\text{series}} = x_A + x_B$ for equal loads) immediately rather than re-deriving the compliance formula from scratch; that recognition alone saves 60 to 90 seconds.

Common traps

Students routinely lose marks on Hooke's law questions for reasons that have nothing to do with arithmetic. The most common is treating Hooke's law as if it applies to any amount of deformation of a solid — it is valid only up to the elastic limit, and every problem statement that mentions "obeys Hooke's law" is implicitly telling you the data lies within that straight-line region. A second, closely related trap is confusing the elastic limit with the fracture point: a material can be stressed well past its elastic limit and still not break, it simply stops returning to its original shape once unloaded (it acquires a permanent set). A third trap is forgetting that the modulus of elasticity has units of stress (pascal) precisely because strain is dimensionless — some students mistakenly try to assign it units of newtons or metres, or forget units altogether when quoting a numerical answer. A fourth trap is assuming a single modulus describes a material completely: Young's modulus applies to longitudinal stretching, bulk modulus to uniform volume change, and shear modulus to a sideways slide — each is a separate "Hooke's constant" for a separate kind of deformation, and a material's Young's modulus tells you nothing directly about its shear modulus without further information.

You should be able to…

State Hooke's law precisely, including the "within the elastic limit" condition, and explain why it fails beyond that limit.
Write $\sigma = Y\varepsilon$ (or $F = k_s x$ for a spring) and correctly identify the modulus of elasticity or force constant as the proportionality constant, with correct SI units.
Derive $Y = FL/(A\,\Delta L)$ from the definitions of stress and strain, reproducing each step.
Calculate a modulus of elasticity, a stress, a strain, or an extension from any three of the four related quantities.
Read a stress-strain graph and identify the region where Hooke's law holds, the elastic limit, the yield point, and the fracture point.
Combine two Hookean springs in series or parallel and predict the combined force constant and extension.

Ask the tutor: Curious why some materials like rubber or bone don't stay on that straight line for long — ask the AI tutor to compare Hookean and non-Hookean stress-strain behaviour with real material examples.

Is Hooke's law only for springs, or does it apply to solid wires and rods too?

It applies to any elastic solid, not just coiled springs. The spring version $F = k_s x$ and the general stress-strain version $\sigma = Y\varepsilon$ are the same statement written in two different sets of variables — a spring is just a convenient shape where the "stress" and "strain" get bundled into force and displacement directly. A copper wire, a steel rod, or a rubber band all obey the same proportionality within their respective elastic limits.

Why does rubber feel like it disobeys Hooke's law even for small stretches?

Rubber's molecular structure is long, coiled polymer chains rather than a rigid crystal lattice, so its stress-strain graph curves almost from the very start instead of staying straight for a good stretch like a metal does. Rubber is still elastic — it snaps back completely when released — but it is not well described by a single constant modulus over any appreciable range, so it is often called a non-Hookean or non-linear elastic material. Metals like steel and copper stay close to Hookean over a much wider range of stress, which is why textbook problems almost always use them.

If I double the load on a wire, does the extension always double?

Only if the resulting stress still lies within the elastic limit — that is exactly what Hooke's law guarantees, since $\Delta L = \sigma L / Y = (F/A)(L/Y)$ is directly proportional to $F$ when $A$ , $L$ , and $Y$ are fixed. But if doubling the load pushes the stress past the elastic limit, the extension will grow faster than proportionally, because the material has left the straight-line region of its stress-strain graph.

How is the elastic potential energy formula connected to Hooke's law?

Because stress rises linearly with strain under Hooke's law, a graph of stress against strain is a straight line through the origin, and the energy stored per unit volume is the area under that line, a triangle: $u = \tfrac12 \times \text{base} \times \text{height} = \tfrac12\,\varepsilon\,\sigma$ . Substituting $\sigma = Y\varepsilon$ gives the equivalent form $u = \tfrac12 Y\varepsilon^2$ . This is exactly analogous to the familiar spring energy $\tfrac12 k_s x^2$ , and it only holds while the material is behaving Hookeanly.