Question

A bungee jumper with mass 63.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 44.0 s . He finally comes to rest 23.0 m below the level of the bridge.Estimate the spring stiffness constant of the bungee cord assuming SHM.

Answer 1

Answer:

80.6 N/m

Explanation:

mass (m) = 63 kg

time (s) = 44 s

number of oscillations (n) = 8

stretched length of the cord (L) =  23 m

we can calculate the spring constant of the cord from the formula below

f =  $\frac{1}{2π}$ x $\sqrt{\frac{k}{m} }$ ...equation 1

where

f = frequency

k = spring constant

m = mass

frequency =$\frac{number of oscillation}{time}$

frequency =$\frac{8}{44}$ = 0.18

now we can input all the required values into the equation 1

0.18 = $\frac{1}{2π}$ x $\sqrt{\frac{k}{63} }$

0.18 x 2π = $\sqrt{\frac{k}{63} }$

1.13 = $\sqrt{\frac{k}{63} }$

$1.13^{2}$ = $\frac{k}{63}$

k = 63 x 1.28 = 80.6 N/m