Theory — Hooke's Law and the Spring Constant
Hooke's Law
When you stretch or compress a spring, it pushes or pulls back. For small stretches the restoring force is proportional to how far the spring is stretched from its natural (unstretched) length. This relationship is Hooke's law, and the constant of proportionality is the spring constant k, which measures how stiff the spring is.
F = force applied to the spring (N)
x = extension from natural length (m)
k = spring constant (N/m) — larger k means a stiffer spring
A Hanging Mass
When a mass hangs at rest from a vertical spring, it is in equilibrium: the upward spring force exactly balances the downward weight of the mass. Setting the spring force equal to the weight lets us connect the stretch to the mass.
k · x = m · g
so x = (g / k) · m and m = k · x / g
Finding k from a Graph
If you hang several known masses and record the extension for each, you can plot the spring force F = mg (vertical axis) against the extension x (horizontal axis). Hooke's law says this graph is a straight line through the origin, and its slope is the spring constant k. Using a graph of several points is more reliable than a single measurement because it averages out the small errors in any one reading.
slope of the line = k
Finding an Unknown Mass
Once the spring constant is known, the same spring becomes a scale. Hang an object of unknown mass, measure how far the spring stretches, and rearrange the equilibrium equation to find the mass.
Stiffer spring
Larger k. For the same mass it stretches less. The F-vs-x graph is steeper.
Heavier mass
Greater weight, so the spring stretches further. Extension ∝ mass.
The graph slope
Plotting F = mg against x gives a straight line whose slope equals k.
| Quantity | Symbol | Relationship |
|---|---|---|
| Spring constant | k | stiffness; slope of F vs x (N/m) |
| Extension | x | stretch from natural length (m) |
| Spring force | F | F = kx (N) |
| Hanging mass equilibrium | — | kx = mg |
| Unknown mass | m | m = kx/g |
Instructions — Running the Virtual Experiment
The Spring Lab tab lets you hang masses and read the extension; the Find k tab records force against extension so you can find the spring constant from the slope and then weigh the unknown objects. Record every reading in your lab notebook.
Simulation — Masses on a Spring
Spring & gravity
Hang a mass
Unknown objects
Record known masses
| Mass (g) | Spring force F = mg (N) | Extension x (m) | F / x (N/m) |
|---|---|---|---|
| No readings yet — click a mass to record it. | |||
Team Questions
Example Lab Report
Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.
The Spring Constant of a Spring
Physics | Section: [Your Section] | Date: [Date]
Lab Members: [Names of all members present]
Purpose
To determine the spring constant of a spring by hanging known masses, measuring the extension produced by each, and finding the spring constant from the slope of a graph of spring force against extension. The measured spring constant is then used to determine the masses of two unknown objects (Red and Blue) from the extensions they produce.
Theory
A spring obeys Hooke's law, F = kx, where k is the spring constant. For a mass hanging in equilibrium, the spring force balances the weight, so kx = mg. A graph of F = mg against extension x is a straight line through the origin with slope k. Once k is known, an unknown mass is found from m = kx/g.
slope of F-vs-x graph = k (g = 9.81 m/s²)
Calculations — Sample data (Earth gravity, middle spring)
For 100 g: F = mg = 0.100 × 9.81 = 0.981 N; measured extension x = 6.54 cm = 0.0654 m
Slope check: k = F/x = 0.981 / 0.0654 = 15.0 N/m
Red unknown: extension 11.45 cm = 0.1145 m → m = kx/g = 15.0 × 0.1145 / 9.81 = 0.175 kg = 175 g
Blue unknown: extension 8.18 cm = 0.0818 m → m = kx/g = 15.0 × 0.0818 / 9.81 = 0.125 kg = 125 g
Results Table — Known Masses
| Mass (g) | Spring force F = mg (N) | Extension x (m) | F / x (N/m) |
|---|---|---|---|
| 50 | 0.491 | 0.0327 | 15.0 |
| 100 | 0.981 | 0.0654 | 15.0 |
| 150 | 1.472 | 0.0981 | 15.0 |
| 200 | 1.962 | 0.1308 | 15.0 |
| 250 | 2.453 | 0.1635 | 15.0 |
A graph of F against x was a straight line through the origin; its slope gave k = 15.0 N/m.
Discussion
The spring force was directly proportional to the extension, confirming Hooke's law over the range of masses tested. The graph of F against x was a straight line through the origin, and its slope gave a spring constant of 15.0 N/m. The constancy of the ratio F/x across all five masses (15.0 N/m in every row) provided an independent confirmation of the slope.
Using this spring constant, the two unknown objects were weighed from their extensions: the Red object (extension 11.45 cm) had a mass of 175 g, and the Blue object (extension 8.18 cm) had a mass of 125 g. The graphical method is more reliable than a single measurement because plotting several points averages out the random error in any individual reading and makes any systematic problem (such as a non-zero starting extension) visible as a departure from a straight line through the origin.
Conclusion
The spring constant of the spring was found to be 15.0 N/m from the slope of the force-versus-extension graph, consistent with the constant F/x ratio. Using this value, the unknown Red and Blue objects were determined to have masses of 175 g and 125 g respectively, demonstrating that a calibrated spring can serve as a scale through Hooke's law.
Practice Questions
Show all work and include units in your answers. Use g = 9.81 m/s² unless told otherwise.