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

80.6 N/m

Step-by-step 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 =
`(1)/(2π)` x
`\sqrt{(k)/(m) }` ...equation 1

where

f = frequency

k = spring constant

m = mass

frequency =
`(number of oscillation)/(time)`

frequency =
`(8)/(44)` = 0.18

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

0.18 =
`(1)/(2π)` x
`\sqrt{(k)/(63) }`

0.18 x 2π =
`\sqrt{(k)/(63) }`

1.13 =
`\sqrt{(k)/(63) }`

`1.13^(2)` =
`(k)/(63)`

k = 63 x 1.28 = 80.6 N/m