Answer:
k/π²
Explanation:
According to Hooke's law, the period (T) of a mass on a spring is equal to 2π times the square root of the ratio of the mass (m) to the spring stiffness (k).
[tex]\Large \text {$ T=2\pi $}\huge \text {$ \sqrt{\frac{m}{k}} $}[/tex]
If we square both sides and rearrange to solve for 4m:
[tex]\Large \text {$ T^2=4\pi^2 $}\huge \text {$ \frac{m}{k} $}\\\\\Large \text {$ 4m= $}\huge \text {$ \frac{k}{\pi^2} $}\Large \text {$ T^2 $}[/tex]
Therefore, if we plot 4m on the y-axis and T² on the x-axis, the slope of the line will be k/π².
