Answer:
[tex]\huge \text {$ \frac{8\pi^3}{\sqrt{m^3k^3}} $}[/tex]
Explanation:
According to Hooke's law, the period (T) of a mass on a spring is equal to 2π times the square root of the ratio of the mass (m) to the spring stiffness (k).
[tex]\Large \text {$ T=2\pi $}\huge \text {$ \sqrt{\frac{m}{k}} $}[/tex]
If we cube both sides and solve for T³/m²:
[tex]\Large \text {$ T^3=8\pi^3 $}\huge \text {$ \sqrt{\frac{m^3}{k^3}} $}\\\\\Large \text {$ T^3=8\pi^3m $}\huge \text {$ \sqrt{\frac{m}{k^3}} $}\\\\\huge \text {$ \frac{T^3}{m^2} $}\Large \text {$ \ =$}\huge \text {$ \frac{8\pi^3m}{m^2} \sqrt{\frac{m}{k^3}} $}\\\\\huge \text {$ \frac{T^3}{m^2} $}\Large \text {$ \ =8\pi^3m$}\huge \text {$ \sqrt{\frac{1}{m^3k^3}} $}\\\\\huge \text {$ \frac{T^3}{m^2} $}\Large \text {$ \ =m $}\huge \text {$ \frac{8\pi^3}{\sqrt{m^3k^3}} $}[/tex]
Therefore, if we plot T³/m² on the y-axis and m on the x-axis, the slope of the line will be [tex]\Large \text {$ \frac{8\pi^3}{\sqrt{m^3k^3}} $}[/tex].
