Answer:
[tex]T\propto \dfrac{1 }{\sqrt{k}} \text{ and } k \propto \dfrac{1}{T^{2}}[/tex]
Explanation:
The period T is inversely proportional to the square root of the spring constant, and
The spring constant is inversely proportional to the square of the period.
It all comes from Hooke's Law.
We usually get the equation for the period, but we can easily rearrange it to get an equation for the spring constant.
[tex]\begin{array}{rcl}T &= &\mathbf{2\pi \sqrt{\dfrac{m}{k}}}\\\\T^{2} &=& 4 \pi^{2} \dfrac{m}{k}\\kT^{2}& =& 4 \pi^{2}m\\k &=& \mathbf{\dfrac{4 \pi^{2}m}{T^{2}}}\end{array}[/tex]
