Topic 2 of 4
Springs and Hooke's law
A spring's extension is its change in length. Within its proportional range, the force needed to stretch it is directly proportional to that extension.
x = loaded length - unloaded length
F = kx
F is the force magnitude in N, x is extension in m and k is the force constant in N/m. A larger k means a greater force is needed for the same extension.
An extended spring exerts a restoring force on its attached load, towards the unstretched position. For a stationary hanging load with no other vertical forces, the upward force from the spring balances the load's downward weight. The load also pulls down on the spring: that force acts on the spring, not on the load. F = kx gives the spring force's magnitude, so its direction must still be shown.
Worked extension
Find the force constant
A spring has unloaded length 16.4 cm and loaded length 20.0 cm under a total applied stretching force of 2.70 N, including any hanger. The spring is within its proportional range.
x = 20.0 - 16.4 = 3.6 cm = 0.036 m.
k = F/x = 2.70/0.036 = 75 N/m.
Using the total length 0.200 m would give the wrong result. The spring responds to extension from its unloaded reference.
Investigate force against extension
- Secure a suitable spring vertically beside a fixed ruler. Use a light pointer and record the unloaded length in the same arrangement.
- Add known forces in suitable steps, staying within the apparatus's intended range. If using masses, include the hanger in the total mass m and calculate its weight W = mg using the stated gravitational field strength g.
- Allow oscillations to settle. Read the pointer with your line of sight perpendicular to the scale, and calculate extension from the unloaded reference.
- Record repeated readings where useful, then plot force vertically against extension horizontally. Include units on both axes.
- Unload in steps and check whether the spring recovers its original length. Preserve the original readings when evaluating the result.
A hanger already stretches the spring. Do not use a reference measured with the hanger attached while calling subsequent values the total force. Either use a genuinely unloaded reference and total forces, or explicitly analyse changes from a stated pre-load.
Measure extension from a genuinely unloaded spring
Use the same fixed upper reference in both views. The unloaded reference contains no hanger or added load. The loaded force includes the hanger's weight; the pointer is treated as negligible. Extension = 20.0 - 16.4 = 3.6 cm = 0.036 m.
A supplied proportional model
The vertical change divided by the horizontal change gives 3.00 N / 0.040 m = 75 N/m. The graph uses extension, not total spring length.
| Force F / N | Length / cm | Extension x / m |
|---|---|---|
| 0 | 16.4 | 0 |
| 0.75 | 17.4 | 0.010 |
| 1.50 | 18.4 | 0.020 |
| 2.25 | 19.4 | 0.030 |
| 3.00 | 20.4 | 0.040 |
For the straight force-against-extension graph, gradient = ΔF/Δx = (3.00 - 0)/(0.040 - 0) = 75 N/m. Plotting extension vertically against force instead gives gradient 1/k, in m/N.
Actual readings may scatter. Use a suitable best-fit line rather than joining each data point. Do not force a zero intercept unless the reference and evidence justify it: a displaced reference can produce a nonzero intercept without changing the spring's true force constant.
Choose a line by inspection
The crosses below are constructed values for a separate force-extension example. They illustrate scatter and a possible reference offset; they are not observations from the earlier exact model. Which candidate describes the whole pattern more suitably?
- Place a ruler through the central trend over the whole plotted range. Rotate and shift it until the vertical deviations on either side are reasonably balanced, with no systematic run from below to above the line. Do not require equal point counts on each side or force the line through the first and last readings.
- A is a suitable by-eye line. B is too flat: the low-extension points are mostly below it and the high-extension point is above it. A by-eye line is an estimate; it need not match one unique calculator regression exactly.
- Choose two widely separated points on the chosen line to calculate its gradient. They need not be original data points. For A, (0, 0.15 N) and (0.040 m, 3.15 N) give (3.15 - 0.15)/0.040 = 75 N/m.
For a straight line, y = mx + c: m is its gradient and c its vertical intercept. Here y is F, the horizontal variable is extension x, and m is a graph coefficient, not mass. Candidate A therefore represents:
The 0.15 N intercept is the line's predicted reading at x = 0. Check the force zero and unloaded reference before interpreting that offset. A nonzero intercept is not by itself evidence that the spring fails Hooke's law. Keep every original point unless a recorded reason justifies excluding one.
Find where the two lines agree
Candidate B is F = (50 N/m)x + 0.90 N. At their intersection, both predict the same force at the same extension:
= (50 N/m)x + 0.90 N
(25 N/m)x = 0.75 N
x = 0.030 m; F = 2.40 N
This crossing lies within the plotted range. Agreement at one point does not make B a good description of the full data pattern, or make the crossing a measured reading. Predict with the chosen line only over a range where the spring model remains supported.
Proportionality and elasticity are different tests
The limit of proportionality marks where force and extension cease to be directly proportional. The elastic limit concerns recovery of the original length after the load is removed. Returning to the original length alone does not prove that the loading graph was a straight line.
- Pointer viewed at an angle
- Parallax changes the reading. Bring the eye level with the pointer and look perpendicular to the ruler scale.
- Pointer still oscillating
- The length is changing during the reading. Allow it to settle before recording a stationary value.
- Ruler or reference has shifted
- Check that the ruler remains fixed and recheck the unloaded reference. Repeating a shifted reading does not repair the reference.
- Different unloading behaviour
- Record it and investigate whether the original length is recovered. Do not assume the linear model continues beyond the tested range.
Estimate and choose a range
Check the scale before measuring
For an assumed 0.3 kg total load and supplied g ≈ 10 N/kg, expect a force around 3 N. If the spring extends roughly 4 cm, estimate k ≈ 3/0.04 ≈ 75 N/m.
A 0 to 1 N force scale would be too small for that load. A suitable 0 to 5 N scale could cover it, provided its resolution and uncertainty meet the task and the spring and stand are rated for the load. These are planning estimates. Use measured values to determine the force constant with a justified precision.