Hooke's law

Before this

Measuring forces — the newton meter · coming soon
Weight and mass · coming soon
Newton's third law

This leads to

Elastic potential energy stored in a spring
Young's modulus
Simple harmonic motion · coming soon

A scale with equal spaces, and why it can be trusted

Look at a spring balance, the kind that hangs from a hook in a market or a laboratory. Its scale has marks for 1 N, 2 N, 3 N and so on, and the marks are equally spaced. The maker did not test every possible load. They tested one or two, and then simply drew equal steps.

Why does that work? It works only if every extra newton stretches the spring by the same amount. If the second newton stretched the spring more than the first, the marks would need to spread out. So the equal spaces are a claim about springs: the stretch grows in proportion to the force.

Now hang something far too heavy on the balance and take it off again. The pointer no longer returns to zero. Every reading the balance gives from now on is wrong. The rule that made the scale possible has a limit, and after the limit the spring has changed.

This lecture is about both halves of that story: the rule, and where it stops. The last topic ended with a promise to turn to materials, starting with springs. Here they are. The energy a stretched spring stores comes in the next lecture.

Measure the stretch, not the length

Hang a spring from a clamp. With nothing on it, it has its natural length. Hang a load on it, and it becomes longer. The increase in length is the extension, written $x$ :

extension x = stretched length - natural length

A spring of natural length 12 cm is stretched to 18 cm by a 3 N load. The extension is the increase, x = 6 cm, not the new length. (Drawing not to scale.)

Here is a real-looking set of measurements for one spring. Loads are added one newton at a time.

Load $F$ in N	Length in cm	Extension $x$ in cm
0	10.0	0
1	12.0	2.0
2	14.0	4.0
3	16.0	6.0
4	18.0	8.0

Look at the extension column. Double the load, and the extension doubles: 1 N gives 2.0 cm, and 2 N gives 4.0 cm. Triple the load, and the extension triples. Now look at the length column. When the load doubles from 1 N to 2 N, the length goes from 12.0 cm to 14.0 cm. That is not double. The length is not proportional to the load. Only the extension is.

Stop and think

A student plots load against length for this spring and gets a straight line. She says, "Straight line, so length is proportional to load." Is she right?

Show the answer

No. A straight line shows proportion only if it passes through the origin. Her line meets the length axis at 10.0 cm, the natural length, because the spring already has a length with no load. Proportional means "zero load, zero stretch", and that is true of the extension, not the length. Subtract the natural length first, and the line moves to pass through the origin.

In the table, each newton adds the same 2.0 cm. So the load divided by the extension is the same in every row: $1 \div 0.020 = 2 \div 0.040 = 50$ newtons per metre. That constant ratio is the whole law.

Hooke's law

Up to the limit of proportionality, the extension of a spring is proportional to the force stretching it:

$F = k x$

$F$ is the force in N, $x$ is the extension in m, and $k$ is the spring constant in N/m.

The spring constant measures stiffness. It is the force needed for each metre of extension. A stiff spring, like one in a car's suspension, has a large $k$ : a big force gives only a small stretch. A soft spring, like the one in a retractable pen, has a small $k$ . The spring in the table has $k = 50$ N/m, so 50 N would stretch it by a whole metre, if it could go that far without changing.

Notice the units. $k$ is in newtons per metre, so $x$ must be in metres when you use $F = k x$ . This is where most marks are lost, as Worked Example 2 shows.

Which force, and which way?

When you pull a spring with a force $F$ , the spring pulls back on you with a force of the same size in the opposite direction. That is Newton's third law. So $F = k x$ gives the size of both forces: the force on the spring, and the force the spring exerts. The spring's own force always points back towards its natural length. Stretch it, and it pulls inwards. Some books write this as $F = - k x$ . The minus sign is not a different law. It only says that the spring's force is opposite to the extension.

The same law usually works for compression too. Push a spring shorter by $x$ , and it pushes back with $k x$ , until its coils touch.

Seeing the law on a graph

Plot force on the vertical axis against extension on the horizontal axis. Hooke's law gives a straight line through the origin, and its gradient is $k$ . A steeper line means a stiffer spring. (Some books plot extension up the side instead. Then the line is the same shape, but its gradient is $1/ k$ . Always check which way round the axes are.)

Real springs obey the law only for a while. Keep adding load and two things happen, one after the other.

Force against extension for an invented spring with k = 50 N/m. The line is straight up to P, the limit of proportionality (6 N, 12 cm). Between P and E, the elastic limit (7 N, 14.4 cm), it curves but the spring still returns to its natural length. After a load of 10 N it is unloaded along the dashed line and is left 6.4 cm longer than before.

P — the limit of proportionality. The last point where the graph is straight. Beyond P, $F = k x$ no longer holds: each extra newton stretches the spring more than the one before.
E — the elastic limit. The largest load from which the spring still returns exactly to its natural length. Beyond E, some of the stretch is permanent.

Below E, the spring behaves elastically: remove the load and it returns to its original shape. Beyond E, it behaves plastically: it stays deformed. When the 10 N load is removed, the spring comes back along a line that is parallel to the original straight part, but it ends 6.4 cm along, not at zero. That 6.4 cm is the permanent extension. This is exactly what ruined the market balance in the opening story.

This is the hard part of the lecture, so spend time here. The two limits are different ideas. P is about the shape of the graph: straight or not. E is about what happens when you let go: back to the start or not. For a steel spring they are very close together, and many books treat them as one point. But they are not the same thing, and an exam may ask about either.

Try it on the spring below. Load it, then take the load off, and watch both the picture and the graph.

Explore — load a spring, then let go

Increase the load and watch the extension and F ÷ x. Then press "Remove the load". Do this once below 6 N, once between 6 N and 7 N, and once above 7 N. When does the spring stop returning to its natural length?

Presets:

Load:

The same invented spring (k = 50 N/m, natural length 10 cm), loaded and then unloaded.

Load in N	Extension in cm	F ÷ x in N/m	After the load is removed
2	4.0	50	Returns to 10 cm
6	12.0	50	Returns to 10 cm
6.5	13.1	≈ 49.6	Returns to 10 cm
7	14.4	≈ 48.6	Returns to 10 cm (the elastic limit)
8	17.6	≈ 45.5	Stays 1.6 cm longer
10	26.4	≈ 37.9	Stays 6.4 cm longer

F ÷ x stays at 50 N/m up to 6 N (Hooke's law holds), then falls. Up to 7 N the spring still returns; beyond 7 N it is permanently stretched.

What should you have seen? Up to 6 N, F ÷ x stays at exactly 50 N/m, and the spring always comes back. Between 6 N and 7 N, F ÷ x starts to fall, so the law has failed, yet the spring still comes back. Above 7 N, it does not come back. Loading it again, it starts from its new, longer length.

Stop and think

A rubber band stretches a long way and returns to its original length when you let go. Does that mean a rubber band obeys Hooke's law?

Show the answer

No. A rubber band is elastic: it returns to its original shape. But its force–extension graph is curved almost from the start, so the extension is not proportional to the force. Elastic means "comes back". Obeying Hooke's law means "straight line through the origin". A material can do the first without the second.

Two springs together

Engineers often join springs. What happens to the stiffness? Make a guess before you read on.

Predict — two springs end to end

One spring stretches 4 cm under a 2 N load. Two identical springs are hooked end to end, and the same 2 N load hangs from the bottom. What is the total extension? Choose, then read why.

Two identical springs, each stretching 4 cm under 2 N.

In series, each spring carries the full 2 N, so each stretches 4 cm: 8 cm in total. In parallel, the springs share the load, 1 N each, so each stretches 2 cm, and the load moves down only 2 cm.

The reason is the force in each spring. In series (end to end), every spring carries the whole load, so the extensions add up. The same force gives twice the extension, so the pair has half the stiffness. In parallel (side by side, joined by a bar), the springs share the load, half each. Each stretches half as much, so the pair is twice as stiff.

For two identical springs, each with spring constant $k$ :

k_{series} = \frac{1}{2} k

k_{parallel} = 2 k

The same reasoning works for springs that are not identical. In parallel, the forces add, so $k = k_{1} + k_{2}$ . In series, the extensions add, so $\frac{1}{k} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$ . You do not need to learn these as new facts. Each one comes from asking: is it the force that is shared, or the extension?

Worked examples

Worked Example 1 — Finding the spring constant

A spring has a natural length of 12 cm. A 3 N force stretches it to 18 cm. Find $k$ .

Extension: $x = 18 - 12 = 6$ cm $= 0.06$ m.

$k = \frac{F}{x} = \frac{3}{0.06} = 50$ N/m.

Check: $50 \times 0.06 = 3$ N ✓. This is the spring in the first figure.

Worked Example 2 — Centimetres into metres

A spring has $k = 200$ N/m. What force gives an extension of 15 cm?

Convert first: 15 cm $= 0.15$ m. Then $F = k x = 200 \times 0.15 = 30$ N.

Check: if you had used 15 instead of 0.15, you would get 3000 N, roughly the weight of four adults. A hand-sized spring that needs 3000 N for a 15 cm stretch is not the spring in the question. Always ask whether the size of the answer makes sense.

Worked Example 3 — Reading a graph

A force–extension graph passes through (2 cm, 4 N), (4 cm, 8 N), (6 cm, 12 N) and (8 cm, 14 N). Find $k$ and say where the limit of proportionality is.

The first three points lie on a straight line through the origin. Its gradient is $\frac{12}{0.06} = 200$ N/m.

From 6 cm to 8 cm the force rises only 2 N, not the 4 N the straight line predicts. So the graph bends at about 6 cm, 12 N. That is the limit of proportionality. The graph alone cannot show the elastic limit: for that you must remove the load and see whether the spring returns.

Check: $200 \times 0.04 = 8$ N ✓.

Worked Example 4 — A hanging mass

A 300 g mass hangs from a spring with $k = 25$ N/m. Find the extension. Take $g = 9.8$ N/kg.

The spring is stretched by the weight, not by the mass. $W = m g = 0.300 \times 9.8 = 2.94$ N.

$x = \frac{F}{k} = \frac{2.94}{25} = 0.1176$ m $\approx 11.8$ cm.

Check: $25 \times 0.1176 = 2.94$ N ✓. A mass in grams must become a weight in newtons before it goes into $F = k x$ .

Worked Example 5 — Series and parallel

Two springs, each with $k = 60$ N/m, carry a 6 N load. Find the extension when they are in series, and when they are in parallel.

Series: $k = 30$ N/m, so $x = 6 \div 30 = 0.2$ m $= 20$ cm. Each spring carries 6 N and stretches 10 cm.

Parallel: $k = 120$ N/m, so $x = 6 \div 120 = 0.05$ m $= 5$ cm. Each spring carries 3 N.

Check: one spring alone would stretch $6 \div 60 = 0.1$ m. Series doubles it to 0.2 m; parallel halves it to 0.05 m ✓.

Three classic mistakes

Trap 1 — Using the length instead of the extension

$x$ is the increase in length. A spring stretched from 12 cm to 18 cm has $x = 6$ cm, not 18 cm. Always subtract the natural length first.

Trap 2 — Leaving x in centimetres

$k$ is in N/m, so $x$ must be in metres. 15 cm is 0.15 m. A mixed-unit answer can be wrong by a factor of 100.

Trap 3 — Treating the two limits as one

The limit of proportionality is where the graph stops being straight. The elastic limit is where the spring stops returning to its natural length. They are often close, but they are different points, and Hooke's law can fail before the spring is damaged.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A spring has $k = 40$ N/m. Find its extension, in cm, under a 6 N load.

Hint 1
Rearrange $F = k x$ to make $x$ the subject.

Hint 2
$x = F \div k$ gives metres. Then convert.

Full answer
$x = 6 \div 40 = 0.15$ m $= 15$ cm. Check: $40 \times 0.15 = 6$ N ✓.
A spring is 20 cm long with no load. A 0.5 kg mass makes it 26 cm long. Find $k$ . Take $g = 9.8$ N/kg.

Hint 1
Find the weight, and find the extension in metres.

Hint 2
$W = 0.5 \times 9.8 = 4.9$ N and $x = 0.06$ m.

Full answer
$k = 4.9 \div 0.06 \approx 81.7$ N/m (≈ 82 N/m to 2 s.f.). Check: $81.7 \times 0.06 \approx 4.9$ N ✓.
Two springs, each with $k = 30$ N/m, hold a 9 N load. Find the extension in parallel and in series.

Hint 1
In parallel, is the force shared or the extension shared?

Hint 2
Parallel: $k = 60$ N/m. Series: $k = 15$ N/m.

Full answer
Parallel: $9 \div 60 = 0.15$ m $= 15$ cm (each spring carries 4.5 N). Series: $9 \div 15 = 0.6$ m $= 60$ cm (each carries 9 N and stretches 30 cm). Check: series is four times parallel, because the stiffness is $\frac{1}{2} k$ instead of $2 k$ ✓. (In practice, check that 9 N is below each spring's limit of proportionality.)
You are designing a spring balance that reads from 0 to 10 N on a scale 20 cm long. What spring constant do you need, and how far apart are the 1 N marks? A student hangs 12 N on it, and afterwards, with no load, the pointer sits at 0.6 N. Explain what happened and what this does to later readings.

Hint 1
10 N must give 20 cm $= 0.20$ m of extension.

Hint 2
A pointer that does not return to zero means the spring has a permanent extension. Which limit was passed?

Full answer
$k = 10 \div 0.20 = 50$ N/m. Each newton gives $1 \div 50 = 0.02$ m, so the marks are 2 cm apart, equally spaced because the extension is proportional to the load. The 12 N load took the spring past its elastic limit, so it is permanently stretched by about 0.6 N's worth of scale, $0.6 \times 2 = 1.2$ cm. Every later reading starts from the wrong place, so it is about 0.6 N too high. If the spring's stiffness has also changed, the errors differ from load to load, and the balance must be recalibrated or replaced. The design must keep the full-scale load, 10 N, below the limit of proportionality, so that the equal marks stay true.

The bigger idea

Robert Hooke first hid his law in a puzzle. In 1676 he printed a scrambled string of letters, an anagram, so that he could later prove he had found it first. In 1678, in his Lectures de Potentia Restitutiva, or Of Spring, he gave the answer in Latin: ut tensio, sic vis, "as the extension, so the force".

The law reaches far beyond springs. Any object that is pushed a little away from a resting position, and pulled back by a force proportional to the distance, behaves like a spring. The bonds between atoms in a solid act like tiny springs for small stretches, which is why a steel wire or a wooden ruler also obeys Hooke's law at first. A later lecture uses this to describe whole materials, not single springs. And a force that pulls back in proportion to the displacement makes things vibrate in a very regular way, the idea behind clocks, musical instruments and the motion of atoms. You will meet that later as simple harmonic motion.

But first, a question this lecture has left open. You did work to stretch the spring. Where did that energy go? The next lecture answers it.

Key concepts in this lesson

The extension of a spring is its stretched length minus its natural length, and it is the extension, not the length, that is proportional to the force.
Hooke's law, F = kx, holds up to the limit of proportionality, with F in N, x in m and the spring constant k in N/m.
The spring constant measures stiffness, and it is the gradient of a graph of force (vertical) against extension (horizontal).
The limit of proportionality is where the graph stops being straight, while the elastic limit is where the spring stops returning to its natural length.
Below the elastic limit a material behaves elastically, and beyond it some of the deformation is permanent, which is plastic behaviour.
Identical springs in series halve the spring constant because each carries the whole load, and in parallel they double it because they share the load.