Question

A bungee jumper, whose mass is 85 kg, jumps from a tall building. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in 6.8 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.

Answer 1

Answer:

K= 290.28 N/m

Explanation:

Given: mass= 85 kg

the time taken to reach point two more times in 6.8 s.

2×t= 6.8 sec

t= 6.8/2= 3.4 sec

then, the time period for oscillation is

$t= 2\pi\sqrt{\frac{m}{k} }$

Here K= spring constant

m= mass of jumper

⇒$K= \frac{4\pi^2m}{t^2}$

now plugging the values we get

$K= \frac{4\pi^2\times85}{3.4^2}$

K= 290.28 N/m