# Hooke's law, finding spring constant

A weight that's attached to a spring is pulled down $6.9$cm from the the postion where the spring is still with the weight on (F=G), and then released. The graph (picture) shows the force F from the spring in newtons as a function of the time in seconds. Use the graph to find the spring constant $k$.

I tried to find the spring constant using Hook's law, $F=kx$. The force used when the spring is extended $6.9$cm is $12$N, so $k=F/x$ should give the answer, but apparently its not the case.

What am i missing here? How should i go about finding the spring constant?

• I've to correct me, see my answer. Commented Apr 10, 2017 at 17:26
• A part of the 12 N comes from gravity. You should use only the part caused by the spring. Commented Apr 11, 2017 at 7:43

The force in the graph is measured relative to the unloaded spring whereas the weight is pulled down 6.9cm relative to the equilibrium position of the loaded spring.

The force used to extend the spring by $\Delta x=x_2-x_1=6.9cm$ is the difference $\Delta F=F_2-F_1$ between the maximum force $F_2$ and the equilibrium (mean) force $F_1=kx_1=mg$, both of which can be found from the graph. The spring constant is then given by $\Delta F=k\Delta x$ because $F_1=kx_1$ and $F_2=kx_2$.

• How come i can use \Delta F=k\Delta x but not the regular Hook's law formula? Using the method you described gave me the right answer, i just don't really understand why.
– Pame
Commented Apr 11, 2017 at 7:39
• @Pame Because the difference $\Delta F$ that Sammy uses here is the force caused by the spring. The rest of the 12 N is caused by gravity, and you don't want to include that in Hooke's law. Commented Apr 11, 2017 at 7:45

By newton's law you will find for the force of the pendulum without friction and gravity:

$$m\ddot{x} = -kx$$

For this you get the solution: $x(t) = A \cdot sin(\omega \cdot t)$. Where $\omega = \sqrt{k/m}$. Now the recorded force is proportional to the $\ddot{x}(t)$. For this you obtain:

$$\ddot{x}(t) = -A \cdot \omega^2 sin(\omega \cdot t)$$

And by reading of the period $T$ you can calculate $\omega$.

The mistake was to ignore that it is a differential equation.

$\textbf{Edit:}$

It is true that $k = -\frac{F}{x(t)} = -\frac{m\ddot{x}(t)}{x(t)}$ in the sense that $x(t)$ is the solution of the differential equation. Of course mathematically you have to be carefull at times $t_0$ where $x(t_0)=0$. And since $x(t)$ is oscillating it is clear that $F(t)$ is oscillating.

Here is an alternative method.

The period $T$ of a spring-mass system is $T = 2\pi \sqrt{\dfrac m k}$.

You can find the period from the graph, so you now need to find the mass $m$.

The mass oscillates about its static equilibrium position and you can find the force exerted by the spring $F_{\text{static equilibrium}}$ at this position from the graph because the oscillations of the mass are symmetrical about this position.

At static equilibrium the magnitude of the force exerted by the spring $F_{\text{static equilibrium}}$ is equal to the magnitude of the weight of the mass.