Answer:
The spring constant of the spring is 10.3 N/m.
Explanation:
Given that,
Mass of a bungee jumper, m = 59 kg
The period of oscillation, T = 0.25 min = 15 sec
We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Where
k is the spring constant
[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]
So, the spring constant of the spring is 10.3 N/m.
