Hooke’s Law states that the extension of a spring is directly proportional to the force applied. I expect this to be shown in my results and graphs.

A force (F) on a spring is linearly dependent on its extension, Δx. Hooke’s law states that to extend a spring by an amount Δx, one needs a force, which is determined by the equation F = kΔx. The spring constant, k, is a quality particular to each spring, usually determined by its thickness and the malleability of the metal (the stretchiness of the spring).

In this experiment, the force is being applied through the addition of weights; therefore the force will act downwards and be a product of a combination of mass and gravity. So a more specific formula for this experiment is:

kΔx = mg

So

Fs – Fg = 0

Fs being the upward pulling force applied by the spring.

Fg being the downward pulling force applied by the weight(s).

When the mass is attached to the spring, the forces Fg and Fs are opposed. The spring will stretch until the upward force equals the downward force.

The upward force increases proportionally with the downward force, as the spring is stretched, until the spring becomes forced beyond its elastic limit, and so is warped out of shape. Some springs take on a new, more elongated shape when this happens, and other, more brittle springs snap. This should be noted for a safer experiment.

For the different arrangements of the springs I predict that the parallel springs will have a smaller extension for the same weights, as this is essentially doubling the spring constant.

The opposite will happen for the two springs in series, as the greater length will halve the spring constant, as the springs are basically forming one, longer spring with the same width, thickness etc. This will result in a doubled result, as the weight is not shared.

I will test these predictions from my graphs; k = 1

Gradient

I will divide my results by 100 to gain k in nm-1 (the results would otherwise be in cm).

Observations/results

Analysis

On the next pages is the graph of my results.

I will find the value of k for all the different spring arrangements;

Parallel: k = Δy = 1.1 = 0.44 = 0.0044 nm-1

Δx 2.5

Single: k = Δy = 0.50 = 0.22 = 0.0022 nm-1

Δx 2.25

Series: k = Δy = 1.0 = 0.14 = 0.0014 nm-1

Δx 7.0

The spring constants for parallel and series are obviously linked. For these values my results are surprisingly accurate, however, for the series line, although the value is much lower than the single line, it is not half.

This is probably due, in fact, to the lower k itself, as this would have caused greater oscillation to the springs when we applied the weights, and so confused our results.

However, mainly, my results support my prediction, although there were a few anomalous points. The reasons for these will be explained in the evaluation section.

My results agree with, prove and support Hooke’s Law.