Question

Check the ongoing discussion below
I'm pretty lost with this one, I think they are asking for the young's modulus. But they have given me data to calulate the period, frequency. So maybe I need to interpret it as a damped osccilator?

A bungee jumper with mass 64.0kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 47.0s . He finally comes to rest 24.0m below the level of the bridge.

Estimate the spring stiffness constant of the bungee cord assuming SHM.

Estimate the unstretched length of the bungee cord assuming SHM.

Answer 1

For this question, you can't assume SHM because that would assume the man has no weight and the bungee cord exerts a push when compressed, because in the definition of SHM force is proportional to displacement from the rest point, which clearly isn't the case here.

Answer 2

l = length of rope
k = spring constant
g = gravitaional acceleration
\[64g=k(l-24)\]
(Weight of jumper is the same as the tension in the cord at the end point.)

Answer 3

ok, so you are stating that the left side, essentially the force due to gravity on the jumper is equal to the spring constant multiplied by the length of the rope - the equilibium point when the jumper is not osscilating. Well at that point the jumper is in equibrium so that makes sense by hookes law. The l-24 is essentially the change in x.

Answer 4

that's good but i cant solve for k if I don't know l in this scenario. Clearly the same applies for calculating l.... unless there is something I'm missing

Answer 5

That's why we need to derive another equation. I'll have a think about it.

Answer 6

ok, im gonna eat lunch but I'll keep an eye on this thread

Answer 7

F = upward force on jumper
d = distance below bridge
So when d>24 then\[F=k(d-l)-64g\]Otherwise the cord is loose so F is just 64g.
By the way, in my earlier post I meant (24-l) rather than (l-24).

Answer 8

Yes I know, we need another equation now.

Answer 9

Where it says that he hit the low point 8 more times in 47; does that mean that the first of those times coincided with the start of those seconds and the eighth with their end?

Answer 10

l is equal to the length of the unweighted cord right? Because as long as d is greater than l then he cord is being stretched. But l is somewhat less than 24m because 24 m is the length of the weighted cord. so between the unstretched length and the weighted equilibrium point, there is still an upward force on the jumper from the cord. Ie between l and 24m the cord is not loose.

Answer 11

I interpreted that as starting from the low point because it says 8 more times. If you draw it out i got 8 osscilations in 47s. for a frequency of .170Hz and a period 5.87 seconds

Answer 12

There is a difference between 8 oscillations and hitting the low point 8 times in 47s. With the latter, only 7 oscillations are possible if the first and last low points coincided with the start and end of that time period respectively.

Answer 13

Well, try drawing it like a -cosine graph. the start at t=0 at the first low point, then add 8 more low points. from that first low point to the last of the additional 8 low points is eight cycles.

Answer 14

I understood that from that low point, 8 low points later totals to 47 seconds

Answer 15

However i suspect that because the bungee jumper eventually stops osscilating then this must be a damped oscillation and thus the formula for the amplitude should be... one sec I'm finding the equation in my textbook