Calculating Bungee Cord Stretch: Mass, Length, and Stiffness Formula

[patrickmoloney]

Homework Statement

A bungee jumper of mass m drops off  a bridge and falls vertically downwards.
The bungee cord is elastic with natural length L and stiffness k. Deduce that
at the lowest point of the fall, the cord is stretched by an amount x = \frac {mg}{k} \Big( 1 + \sqrt {1+ \frac {2kL}{mg}} \Big)

Homework Equations

mgh = \frac{1}{2} k x^2

The Attempt at a Solution

Energy considerations dictate that the gravitational potential energy of the jumper in the initial state is equal to the elastic potential of the cord in the final state

mg ( L + x ) = \frac{1}{2} k x^2

mgL + mgx = \frac{1}{2} k x^2

2mgL + 2mgx = k x^2

x^2 - \Big( \frac{2mg}{k} \Big )x - \Big( \frac{2mgL}{k} \Big) = 0

x = \frac{- \Big(- \frac{2mg}{k} \Big) + \sqrt{ \Big( - \frac{2mg}{k} \Big)^2 - 4 \Big( - \frac{2mgL}{k} \Big )}}{2}

x = \frac{1}{2} \Big( \frac{2mg}{k} + \sqrt{ \frac{4m^2g^2}{k^2} + \frac{8mgL}{k}} \Big)

x = \frac{mg}{k} + \frac{1}{2} \Big ( \sqrt{ \frac{4m^2g^2}{k^2} \Big ( 1 + \frac{2kL}{mg} \Big )} \Big )

x = \frac{mg}{k} + \frac{1}{2} \Big( \sqrt{\frac{4m^2g^2}{k^2}} \sqrt{1+\frac{2kL}{mg}} \Big )

x = \frac{mg}{k} + \frac{1}{2} \Big( \frac{2mg}{k} \Big) \sqrt{1 + \frac{2kL}{mg}}

x = \frac{mg}{k} + \frac{mg}{k} \sqrt{1 + \frac{2kL}{mg}}

x = \frac{mg}{k} \Big( 1 + \sqrt{1+ \frac{2kL}{mg}} \Big)

Is this the correct method? There is no solution online to this problem.

 

[CWatters]

Seems reasonable to me.