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 $k$. Graph

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?

  • $\begingroup$ I've to correct me, see my answer. $\endgroup$
    – Alpha001
    Commented Apr 10, 2017 at 17:26
  • $\begingroup$ A part of the 12 N comes from gravity. You should use only the part caused by the spring. $\endgroup$
    – Steeven
    Commented Apr 11, 2017 at 7:43

3 Answers 3


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.

![enter image description here

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 $F_2=kx_2$.

  • $\begingroup$ 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. $\endgroup$
    – Pame
    Commented Apr 11, 2017 at 7:39
  • $\begingroup$ @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. $\endgroup$
    – Steeven
    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.


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.


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