Before you start
You need these ideas and skills for this practical. If any of them are new, read this section first.
Key words
- Load: the force (weight) stretching the spring, in newtons (N). A 100 g mass gives a load of about 1.0 N.
- Extension (x): the increase in length of the spring: stretched length − original length.
- Elastic: a material is elastic if it returns to its original shape when the force is removed.
- Limit of proportionality: the point beyond which the extension is no longer proportional to the load. The graph stops being a straight line.
- Spring constant (k): the force needed per unit extension: k = F ÷ x. A stiffer spring has a bigger k.
- Anomaly: a result that does not fit the pattern of the others.
- Oscillate: to move repeatedly up and down, as the spring does when a mass is added.
Extension, not length
The key quantity is the extension, not the length of the spring:
If you double the load, the spring does not become twice as long: it is the extension that doubles.
In this practical you never need to measure the spring itself. The zero reading on the rule (hanger only) is your starting point, so:
Reading the rule
The rule is fixed vertically beside the spring with 0 cm at the top, so the readings get bigger as the spring stretches. Read the rule where the pointer on the hanger meets it.
- Keep your eye level with the pointer (no parallax error).
- Read to the nearest millimetre (0.1 cm).
- Wait for the spring to stop bouncing before you read.
- Always read at the same point (the pointer), so every reading is consistent.
Skills you need
- Converting mass to weight: W = mg; 100 g = 0.1 kg gives about 1.0 N.
- Converting units: 1 cm = 0.01 m, so 0.25 N/cm = 25 N/m.
- Recording results in a table with units in the headings.
- Plotting load against extension, drawing a best-fit line, and finding its gradient.
Aim and apparatus
Aim
To find how the load on a spring affects its extension, to determine the spring constant k = F ÷ x from a load–extension graph, and to find the limit of proportionality.
Syllabus link: Cambridge IGCSE Physics 0625 / 0972, 1.5.1 Effects of forces: know that forces may change the size and shape of an object; plot, draw and interpret load–extension graphs for an elastic solid and describe the associated experimental procedures. Supplement: define the spring constant, recall and use k = F/x, and recognise the significance of the limit of proportionality.
Apparatus
- Clamp stand, boss and clamp, and a G-clamp to fix the stand to the bench
- Spring (and a rubber band to compare)
- Hanger with ten 100 g slotted masses
- Metre rule, clamped vertically with 0 cm at the top, and a pointer on the hanger
- Plumb line (a thread with a weight), to check the rule is vertical
- Eye protection, and a mat on the floor under the masses
Theory
Hooke's law
The extension of a spring is proportional to the load on it, provided the limit of proportionality is not exceeded:
F = load in N, x = extension in cm or m, k = spring constant in N/cm or N/m. Robert Hooke published this in 1676 as a coded anagram, ceiiinosssttuv: “Ut tensio, sic vis”, which means “as the extension, so the force”.
The load–extension graph
Plot load on the y-axis and extension on the x-axis (as in your coursebook).
- Up to the limit of proportionality: a straight line through the origin.
- Its gradient is the spring constant: k = ΔF ÷ Δx. A stiffer spring has a steeper line.
- Past the limit of proportionality the line curves: each extra newton stretches the spring more than before.
Past the limit
If a spring is stretched beyond its elastic limit (just past the limit of proportionality), it is permanently deformed. When the load is removed it does not go back to its original length: the pointer stays lower than the zero reading.
A rubber band is different: its load–extension graph is S-shaped, never a straight line, so it does not obey Hooke's law. When unloaded it does not come back exactly along the same curve.
Two springs (extension)
In series (one below the other): each spring carries the full load and stretches by the full amount, so the total extension is doubled and k is halved.
In parallel (side by side): each spring carries half the load, so the extension is halved and k is doubled.
(This goes beyond the IGCSE syllabus, but it is a good test of your understanding of F = kx.)
Planning the investigation
Variables
| Type | Variable | How |
|---|---|---|
| Independent | Load F | Add 100 g (1.0 N) at a time, from 0 to 10 N |
| Dependent | Extension x | Rule reading at the pointer − zero reading |
| Control | The spring | Same spring throughout; do not overstretch it before you start |
| Control | Reading position | Always read at the pointer, eye level, after the bouncing stops |
Getting started (questions from the books)
- Why the G-clamp? To stop the stand toppling over when the masses are added.
- Why the plumb line? To check the rule is vertical, parallel to the spring.
- Why measure from the same place every time? So that every reading has the same starting point and the extensions are consistent. A pointer fixed to the hanger makes this easy.
- What about the bouncing? Wait until the spring is still, or steady it gently with your hand, before reading.
Method in a real lab
- Set up the clamp stand and fix it to the bench with a G-clamp. Hang the spring from the clamp. Put a mat on the floor beneath it and wear eye protection.
- Clamp the metre rule vertically beside the spring with 0 cm at the top. Check it is vertical with a plumb line.
- Hang the hanger on the spring. When it is still, record the reading on the rule at the pointer: this is the zero reading (load 0 N).
- Add a 100 g mass (1.0 N). Wait for the bouncing to stop and record the new reading, with your eye level with the pointer.
- Remove the mass and check whether the pointer returns to the zero reading (is the spring still elastic?).
- Repeat, adding 100 g each time, up to 1.0 kg (10 N) or until the spring is clearly damaged.
- Calculate each extension: reading − zero reading.
- Plot load (y-axis) against extension (x-axis). Draw the best-fit line, circle any anomalies, find the gradient (k) and mark the limit of proportionality.
How to use this virtual lab
- Choose the soft spring, the stiff spring or the rubber band at the top of the panel.
- Step 1, Set up: check the rule with the plumb line, then hang the hanger on the spring.
- Step 2, Zero: the magnifier shows the rule beside the pointer. Read it to 0.1 cm with your eye Level, type it and press Save zero reading.
- Step 3, Load: press + Add 100 g. The spring bounces: wait, or press Steady the spring. Read the pointer, type the reading and press Record reading. Repeat up to 10 masses.
- Unload test: takes all the masses off. Does the pointer go back to the zero reading?
- Step 4, Find k: the lab shows the straight part and the limit of proportionality. Calculate k = F ÷ x yourself and press Check.
- Table & graph: the load–extension graph. Best-fit line draws the straight part and marks the limit of proportionality; Join points joins your own points. Each spring gets its own colour.
- More to explore: two springs in series and in parallel, and a mystery object to weigh with your spring.
Results
Sample results: soft spring
Zero reading (hanger only) = 30.2 cm.
| Load F / N | Reading / cm | Extension x / cm | F ÷ x / N/cm |
|---|---|---|---|
| 0.0 | 30.2 | 0.0 | – |
| 1.0 | 34.2 | 4.0 | 0.250 |
| 2.0 | 38.2 | 8.0 | 0.250 |
| 3.0 | 42.2 | 12.0 | 0.250 |
| 4.0 | 46.2 | 16.0 | 0.250 |
| 5.0 | 50.2 | 20.0 | 0.250 |
| 6.0 | 54.2 | 24.0 | 0.250 |
| 7.0 | 59.6 | 29.4 | 0.238 |
| 8.0 | 67.8 | 37.6 | 0.213 |
| 9.0 | 78.8 | 48.6 | 0.185 |
| 10.0 | 92.6 | 62.4 | 0.160 |
Up to 6.0 N the extension goes up in equal steps of 4.0 cm. After that the steps get bigger.
Calculating k
Use a point on the straight part, or better, the gradient of the best-fit line:
The limit of proportionality is at about 6 N. After unloading from 10 N the pointer stayed several centimetres below the zero reading: the spring was permanently stretched.
Analysis and conclusion
Describing the graph
“The graph is a straight line through the origin up to a load of about 6 N, so the extension is directly proportional to the load (Hooke's law). Above 6 N the line curves towards the extension axis: the spring stretches more for each extra newton. This is beyond the limit of proportionality.”
Comparing springs
The stiff spring's line is steeper (k ≈ 0.60 N/cm) than the soft spring's (k ≈ 0.25 N/cm): it needs a bigger force for the same extension. The rubber band's graph is curved, so it has no single spring constant.
Using a spring as a newton meter
Once k is known, a spring can measure an unknown force: measure the extension and use W = kx. This is how a newton meter (spring balance) works, and why Hooke was so interested in springs. It only works on the straight part of the graph.
Conclusion
“The soft spring obeys Hooke's law up to a load of 6 N. Its spring constant, from the gradient of the straight part, is 0.25 N/cm (25 N/m). Beyond 6 N the extension increases faster than the load, and the spring was permanently deformed.”
Evaluation
| Source of error | Effect | How to reduce it |
|---|---|---|
| Parallax: eye not level with the pointer | Readings too big (eye above) or too small (eye below) | Keep your eye level with the pointer; use a pointer that touches the rule. |
| Reading while the spring is still bouncing | Random errors and anomalous points | Wait until the spring is still, or steady it gently by hand. |
| Rule not vertical (not parallel to the spring) | Readings are not true vertical distances | Check with a plumb line or a set square against the bench. |
| Not reading from the same point each time | Inconsistent extensions | Fix a pointer to the bottom of the hanger and always read there. |
| Rule resolution (1 mm) | Each extension uncertain by about ±0.1 cm | Use large extensions; repeat and average. |
| Spring already overstretched | Zero reading changes; graph does not pass through the origin | Do the unload test; use a new spring if it has been damaged. |
| Masses not exactly 100 g, or using g = 10 N/kg | Small systematic error in the loads | Weigh the masses; use W = mg with g = 9.8 N/kg for accurate work. |
Improvement: take readings while unloading too (removing the masses one at a time) and average them with the loading readings.
Safety
- Fix the clamp stand to the bench with a G-clamp so it cannot topple.
- Wear eye protection: a spring can snap back if it breaks under tension.
- Put a mat on the floor under the masses, and do not stand on it or put your feet under the masses.
- Wear closed shoes in case a mass falls.
Exam practice
Practical-style questions like those in Paper 5 (Practical Test) and Paper 6 (Alternative to Practical), and questions from your coursebook and practical book. Try each one before opening the answer.
1. A piece of elastic cord is 75 cm long. When it is stretched its length is 97 cm. What is its extension?
Extension = 97 − 75 = 22 cm.
2. A spring has a spring constant of 20 N/cm. What load produces an extension of 2.5 cm?
F = kx = 20 × 2.5 = 50 N.
3. A spring needs a load of 7.5 N to extend by 5.0 cm. What load gives an extension of 12 cm?
k = 7.5 ÷ 5.0 = 1.5 N/cm. F = kx = 1.5 × 12 = 18 N (assuming the limit of proportionality is not exceeded).
4. A spring has an unstretched length of 13.0 cm and k = 7.0 N/cm. What load stretches it to 18.0 cm?
Extension = 18.0 − 13.0 = 5.0 cm. F = kx = 7.0 × 5.0 = 35 N.
5. A student hangs ten 0.1 N weights on a spring. With no load the spring is 0.0400 m long; with 0.6 N it is 0.0445 m long. Calculate the extension and the spring constant.
x = 0.0445 − 0.0400 = 0.0045 m. k = F ÷ x = 0.6 ÷ 0.0045 = 133 N/m.
6. In question 5, the extensions for 1.0 N and 1.2 N are 0.0080 m and 0.0105 m. Do the results support “load is directly proportional to extension”?
Up to 0.8 N, yes: equal steps of load give equal steps of extension (0.0015 m per 0.2 N), and the line passes through the origin. At 1.0 N and 1.2 N the steps get bigger (0.0020 m, 0.0025 m), so the spring has gone past its limit of proportionality and the statement no longer holds there.
7. Describe how the student should make sure the rule is parallel to the spring.
Use a plumb line (a thread with a weight) beside the rule to check that the rule is vertical; or use a set square on the bench. Clamp the rule close to the spring.
8. Why does the student wait before taking each reading?
When a mass is added, the spring oscillates (bounces). Reading while it is moving gives inaccurate, anomalous results; the student must wait until it is still.
9. Describe how the graph would change if more and more masses were added, and explain why.
The graph would curve (bend towards the extension axis) after the limit of proportionality: each extra newton gives a bigger extension. The spring has been stretched beyond its elastic limit, so it is permanently deformed and will not return to its original length.
10. Two identical springs, each with k = 25 N/m, are hung side by side to share a load of 10 N. What is the extension?
Each spring carries 5 N, so x = F ÷ k = 5 ÷ 25 = 0.20 m (20 cm). Together they act like one spring with k = 50 N/m.