Problem — Hooke's Law from Experimental Data

Problem

In an experiment on a spring, the following (applied force, extension) pairs are measured: $(1{,}0;\,4{,}5)$, $(2{,}0;\,7{,}0)$, $(3{,}0;\,9{,}5)$, $(4{,}0;\,12{,}0)$ in N and cm. Plot $F$ as a function of the extension $\Delta x$, check whether Hooke’s law holds, and derive the spring constant $k$. Does the graph pass through the origin?

Solution

Step 1 — Organising the data.

$F$ [N]	$\Delta x$ [cm]
1,0	4,5
2,0	7,0
3,0	9,5
4,0	12,0

Step 2 — Checking linearity. Going from one row to the next, the force always increases by $1{,}0\;\text{N}$ and the extension always increases by $2{,}5\;\text{cm}$:

$4{,}5 \to 7{,}0 \to 9{,}5 \to 12{,}0 \quad(\text{constant step } +2{,}5\;\text{cm})$

The increments are constant: the points are collinear, so the relation between $F$ and $\Delta x$ is linear and Hooke’s law is verified (over the range explored).

Step 3 — Spring constant. The slope of the $F$–$\Delta x$ graph is the ratio between the increase in force and the increase in extension. Converting centimetres to metres ($2{,}5\;\text{cm} = 0{,}025\;\text{m}$):

$k = \frac{\Delta F}{\Delta(\Delta x)} = \frac{1{,}0\;\text{N}}{0{,}025\;\text{m}} = 40\;\text{N/m}$

$\ev{k \approx 40\;\mathrm{N/m}}$

Step 4 — Does the line pass through the origin? Extrapolating the line to $F = 0$: the relation reads $\Delta x = 2{,}5\,F + 2{,}0$ (with $\Delta x$ in cm), so for $F = 0$ we would get $\Delta x = 2{,}0\;\text{cm}$. The line does not pass through the origin: it intersects the extension axis at

$\ev{x_0 \approx 2{,}0\;\mathrm{cm}}$

There is, in other words, an initial extension (an offset) already present at zero force, for example due to a preload on the spring or an initial play in the measuring system. The correct law is then $F = k(\Delta x - x_0)$: it remains linear, but is “shifted” relative to the origin.

Links

Topics: Dynamics Concepts: Elastic force and Hooke’s law Skills: Reading graphs Objects: Spring