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?
Best Answer
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 $F_2=kx_2$.