Deformation and Elasticity

1.23 · Hooke’s law

Force, extension, and the linear region

Extension is not the new length. It is how much longer the object has become:

If a spring is $12.0\,\text{cm}$ long with no load and $15.5\,\text{cm}$ with a load, $e = 3.5\,\text{cm} = 0.035\,\text{m}$. Always convert to metres if you are going to find $k$ in N/m.

Hooke’s law says that, in the initial linear region of a force–extension graph, force is proportional to extension:

$k$ is the spring constant (or stiffness). A large $k$ means a stiff spring: you need a big force for a small extension. Unit of $k$ is N/m, because $k = F/e$.

On a graph of force (vertical) against extension (horizontal), the linear region is a straight line through the origin. The gradient of that straight section is $k$:

Draw a large triangle on the straight part only. Do not include the curved part. Do not use a single point if the line does not go through the origin — use the gradient of the straight section you actually drew.

Limit of proportionality and elastic limit

In this course the two ideas sit very close together, and exam mark schemes often accept either phrase.

• Limit of proportionality: the last point where $F$ is still proportional to $e$. After this the graph curves.
• Elastic limit: beyond this, the object will not return to its original length when you unload it. The deformation has become plastic.

For a typical steel spring in a school lab, those two points are near each other. For the exam: if the graph is no longer a straight line, Hooke’s law is no longer obeyed. If the spring is longer after you remove the masses, it has gone past the elastic limit.

1.22 · Required practical

Investigate extension against force for a spring, a wire and a rubber band

The specification names three objects on purpose. The method is the same idea; the graphs are not. I expect a full write-up for the spring, then a paragraph on how the other two differ.

Apparatus

• Retort stand, boss and clamp
• Helical spring (then later: long thin metal wire; rubber band)
• Metre rule, set vertically, and a pointer or paper marker on the bottom of the spring
• Slotted masses and a mass hanger (typically $100\,\text{g}$ steps)
• G-clamp to hold the stand to the bench
• Safety: eye protection, a padded tray or sand tray under the hanging masses

Variables

• Independent: applied force. You control this by adding known masses. Force is the weight, $W = mg$. With $g = 10\,\text{N/kg}$, each $100\,\text{g}$ ($0.100\,\text{kg}$) is $1.0\,\text{N}$.
• Dependent: extension of the specimen.
• Control: same spring/wire/band, same temperature, same measuring instrument, same way of reading the rule.

Method — the version that scores 3 or 4 marks

1. Clamp the stand to the bench. Hang the spring from the clamp. Fix a pointer at the bottom coil. Place the metre rule as close as possible, vertical, with the zero (or a chosen mark) at eye level with the pointer.
2. Record the original length with the hanger only, or with no load — be consistent and state which. This is $L_0$.
3. Add a $100\,\text{g}$ mass. Wait a moment for the spring to settle. Read the new length $L$ at eye level (avoids parallax). Extension $e = L - L_0$. Force $F = mg$.
4. Repeat for at least six forces, in even steps. Record results in a table: mass / kg, force / N, length / m, extension / m.
5. Unload the masses one by one and record the length again. If the lengths match the loading values, the spring was elastic over that range.
6. Plot force on the vertical axis against extension on the horizontal axis (follow the axis the question names). Draw a line of best fit. The straight section through the origin is Hooke’s law; its gradient is $k$.

A results table I would accept

Up to $3.0\,\text{N}$ the extension steps are $2.0\,\text{cm}$ each — linear, $k = 3.0/0.060 = 50\,\text{N/m}$. The $4.0\,\text{N}$ row has jumped more than $2.0\,\text{cm}$. I would say the spring has gone past the limit of proportionality. I would not use that last point to calculate $k$.

How the metal wire is different

A short thick wire barely extends. In school we use a long, thin wire (often a metre or more, small diameter) so the extension is large enough to measure with a ruler or Vernier scale. Clamp one end firmly. Hang masses from the other end, or use a Searle’s apparatus if you have it. The force–extension graph still has a straight Hooke’s-law region, then it may curve. Metal wires can show a clear elastic limit and then plastic flow — the wire stays longer when unloaded. Safety matters more: a snapping wire can whip. Wear eye protection; stand to the side; put a catch box under the masses.

How the rubber band is different

Rubber does not usually give a straight line from the origin. The band is easy to stretch at first, then it gets stiffer, so the graph curves. If you load and then unload, the unloading curve often sits below the loading curve — a loop called hysteresis. You are not required to name hysteresis, but you should be able to say:

• the same method (measure original length, add force, measure extension);
• the graph is non-linear;
• rubber can still be elastic: it may return to its original length when the force is removed, even though $F$ was never proportional to $e$.

That last point is important. Hooke’s law is about the linear region. Elastic behaviour is about recovering shape. A rubber band can be elastic without obeying Hooke’s law.

Errors and improvements

1.24

Elastic behaviour — learn the specification sentence

1.24 is almost a dictation mark. The specification says:

Elastic behaviour is the ability of a material to recover its original shape after the forces causing deformation have been removed.

Write that. Do not replace “shape” with a vague “it goes back.” Do mention that the force has been removed.

• Elastic deformation: remove the load → original length / shape returns.
• Plastic deformation: remove the load → it stays permanently stretched or bent. A bent paperclip is the demonstration I use.

Check you can

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