3.4.1 - Springs and Hooke's law

Deformation, extension and compression

Deformation means a change in shape or size caused by forces. In this topic, the important change is along the length of an object such as a spring, wire, rubber band or strip.

Extension and compression are both changes in length. They are not the final length of the object.

If the original length is $L_0$ and the stretched length is $L$, then:

where $x$ is the extension. For compression, $x$ is the amount by which the length has decreased.

Worked example: finding an extension

A spring has original length $18.0\,\mathrm{cm}$. When a load is attached, its length is $24.5\,\mathrm{cm}$.

Extension:

Convert to metres before using it in $F = kx$:

A common error is to substitute $24.5\,\mathrm{cm}$ as $x$. That is the loaded length, not the extension.

[DIAGRAM: asset_name: 3.4.1 - Springs and Hooke's law — figure 1; asset_slug: 3_04_1_springs_hookes_law_and_force_extension_graphs__diagram_01; recommended_method: drawn_physics; caption: Tension lengthens a specimen while compression shortens it.; description: Outward arrows show tension and inward arrows show compression; initial and loaded spring lengths distinguish length from extension.]

Hooke's law and force constant

Hooke's law is a proportionality law. It applies only over the range where force and extension, or force and compression, are directly proportional.

In this lesson, use the magnitude form:

Here $F$ is the applied force or load in newtons, $x$ is the extension or compression in metres, and $k$ is the force constant in newtons per metre, $\mathrm{N} \mathrm{m}^{-1}$.

The force constant is not a pure material property. A short thick wire and a long thin wire made from the same material can have different force constants because their dimensions are different.

Worked example: calculating a force constant

A spring extends by $3.0\,\mathrm{cm}$ when a load of $2.4\,\mathrm{N}$ is attached.

First convert the extension:

Rearrange Hooke's law:

Substitute:

So this spring needs $80\,\mathrm{N}$ for each metre of extension, or $0.80\,\mathrm{N}$ for each centimetre of extension while it remains Hookean.

Worked example: using $k$ to find extension

A wire has force constant $k = 1.5 \times 10^3\ \text{N m}^{-1}$. A force of $12\,\mathrm{N}$ is applied within the Hookean range.

That is $8.0\,\mathrm{mm}$. Notice that the calculation used metres first; the conversion to millimetres came at the end.

Force-extension graphs

A force-extension graph shows how the applied force changes with extension. For this topic, plot force $F$ on the vertical axis and extension $x$ on the horizontal axis. Then the gradient of the straight-line section is the force constant:

[DIAGRAM: asset_name: 3.4.1 - Springs and Hooke's law - diagram 1; asset_slug: 3_04_1_springs_hookes_law_and_force_extension_graphs__diagram_02; recommended_method: drawn_physics; caption: A straight force-extension line has constant stiffness while the curved response does not.; description: A straight force-extension line of gradient 80 N per metre is compared with a nonlinear response; a gradient triangle shows force change over extension change.]

For a spring or wire obeying Hooke's law:

• the graph is a straight line
• the straight line passes through the origin
• doubling the extension doubles the force
• the gradient is constant, so $k$ is constant

For a rubber band or polythene strip, the graph often curves. A curved graph means the gradient changes, so there is no single constant $k$ for the whole range.

Be careful with axes. If a question gives a graph of extension against force, the gradient is $\Delta x / \Delta F$, so the force constant is the reciprocal of that gradient. The cleanest route for this lesson is usually $F$ against $x$.

Worked example: finding $k$ from a graph

A force-extension graph for a spring has a straight-line region. Two points on the best-fit line are:

• $x = 0.010\ \text{m}$, $F = 0.80\ \text{N}$
• $x = 0.050\ \text{m}$, $F = 4.00\ \text{N}$

Find the gradient:

Use points far apart on the line, not two cramped plotted points. A large gradient triangle reduces the effect of reading uncertainty.

This lesson uses force-extension graphs for shape, proportionality and gradient. The area under a force-extension graph is a later energy idea, so do not use area when the question is only asking whether Hooke's law applies or what $k$ is.