Thread: Converting 2nd order DEs to 1st order

So I have a question that asks to convert the 2nd order equation

$\displaystyle mx'' = mg + b(x) - ax'$

Into a 1st order equation. This equation describes a bungee jump where b(x) is a function for the "spring force" that is 0 when the jumper is above the cord equilibrium and -kx when the jumper is below it. To solve:

I let u1 = x
and u2 = x'

Then:
u1' = x' = u2
u2' = x'' = g + b(x)/m - (a/m)x' = x'' = g + b(x)/m - (a/m)u1

So that's my system. Now, the next question asks me to "Find the distance below the bridge that your feet find themselves when you come to rest after the jump for any given cord with stiffness k. Use the knowledge that you are six feet tall and weigh 160 lbs., and the fact that you do not want to rest below the level of the creek, to specify a lower bound on the value for k that you require for your bungee cord."

At this point I'm a little lost. I'm thinking that I need to solve for x in terms of k using the given constants and then set x = (creek distance) to find the lowest possible k. But do I need to solve the actuall diff eq to do this? And what does the 1st order system of equations have to do with it? We haven't even covered solving system of 1st order equations in class yet, and the mg term in the original differential equation is throwing me for a loop otherwise I would set mx'' + ax' - b(x) = 0 and then use the characteristic equation to solve it easily. The mg is a constant and involves no orders of x so I don't know where to put it.

Also, I'm thrown off by the b(x) function as well, since it can either be 0 or -kx depending on if you're above or below the cord equilibrium. I guess I could solve two differential equations, one for each case, but they alternate back and forth with the jumper oscillating up and down and I really don't know how to incorporate it.

I've probably confused any readers at this point lol. My main goal is just to answer the question posed that I quoted. I don't know where to start. Anyway if anyone could help out that would be amazing. Thanks.

Originally Posted by griffsterb

So I have a question that asks to convert the 2nd order equation

$\displaystyle mx'' = mg + b(x) - ax'$

Into a 1st order equation. This equation describes a bungee jump where b(x) is a function for the "spring force" that is 0 when the jumper is above the cord equilibrium and -kx when the jumper is below it. To solve:

I let u1 = x
and u2 = x'

Then:
u1' = x' = u2
u2' = x'' = g + b(x)/m - (a/m)x' = x'' = g + b(x)/m - (a/m)u1

$\displaystyle \bold{u}=\left[ x \atop x' \right]$

$\displaystyle \bold{u}'=\left[ x' \atop x'' \right]=\left[ u_2 \atop g+\frac{b(u_1)}{m}-\frac{a}{m}u_2 \right]$

Now you come to rest when $\displaystyle \bold{u'}=0$ so you need to solve:

$\displaystyle g+\frac{b(u_1)}{m}=0$


to find the final height.

(note even if your head is above the water at the end does not mean that you will not have been dunked while the system bounces you up and down)

RonL