If you have a linear force-extension graph for say a spring then the spring constant is simply the gradient of the graph. However, how would you calculate the spring constant at a particular point on a non-linear (curved) graph for say an elastic band? Let's imagine you wanted the spring constant at 4cm of extension which corresponds to a load force of 4N. Would you simply do the ratio of the force and extension (i.e. one divided by the other) so 1N/cm in this case or would you find the gradient at that point (i.e. gradient of the tangent at that point)?
[Physics] Spring Constant from a Non-Linear Force-Extension Graph
elasticityspring
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I think the simplest thing is to calculate the energy directly from the data given - using simple numerical integration of the data corresponding to the configuration you are actually using (I am assuming that's the angular configuration).
First, a graph of your data:
The data is quite linear but with a fair bit of noise on it. The best fit quadratic through the "angular" data, forced to go through (0,0), has a small quadratic component. If you just integrated that expression, you would have the expression for energy as a function of displacement:
$$E = \int F\cdot dx = \int (38.5 x^2 + 27.7 x) dx = 12.8 x^3 + 14.4 x^2 + C$$
As long as $x$ is quite small, the cubic term will be small compared to the linear term.
Alternatively, you can use the linear fit through the data - this gives a slope (spring constant) of 32.7 and you would ignore the cubic term.
Here is the calculated energy as a function of displacement using the two methods (over the range 0 - 0.16 units; note that you really should always quote your units when describing an experimental setup...)
As you can see, over this range the linear assumption gives almost the same result.
This is not a complete answer to your question - but I hope it helps your thought process.
For a great explanation you can look on Hyperphysics.
But quickly, you should take the integral $\int_0^x kx'\,\mathrm{d}x'$. This gives you $U=\frac{1}{2} kx^2$.
Best Answer
If you know how force $F$ varies with displacement $x$, $F(x)$, the derivative $\frac{dF(x)}{dx}$ will give you the function $k(x)$.
Hope this helps