Final amplitude with damping of bungee cord

[timetraveller123]

Homework Statement

5. In the spectacular sport of bungee jumping,, a light elastic cord (a bungee cord) is tied tightly around the ankles of someone who jumps from a bridge of height H to which the other end of the cord is attached. The length of the cord is calculated so that the jumper, of mass m, will not quite reach the surface of the water below the bridge before he or she springs back up. Suppose the cord behaves like a spring of spring constant 10 mg/H where g is the acceleration due to gravity.

(a) How long must the cord be so that a jumper just touches the water before being pulled back up? Neglect the height of the jumper and any effects due to friction.

(b) Friction damps the up-and-down motion of the jumper that results after the initial jump. How far above the water would the jumper be when the oscillations have ceased?

Express your answers in terms of
H

so this is the question i have gotten the answer to the first part and stuck at the second part . i could at least solve it if i knew the damping factor or the coefficient of friction but the question gives nothing is there any method i can use to solve this

the given solution is (1/√5 - 1/10)H

Homework Equations

no idea which equation to use

The Attempt at a Solution

 

[BvU]

Hi,

You realize this post doesn't adhere to the PF rules ? 'No idea' and no attempt are real disqualifiers: they make it difficult for potential helpers to distinguish how they can help you effectively.

As a compromise, I propose you post your working for part a). The L you find there will be needed anyway for part b).
And as a hint: 'final amplitude' isn't really a good title (final amplitude is zero: the oscillations have ceased). What you want is the final stretch of the chord in order to add it to L.

 

[timetraveller123]

ah yes sir forgot about the rules and good to see you again hope you remember me when i meant the final amplitude i meant the final stretch

my working for part A:
mgH = 1/2 k x^2

where x = H - L
L is the natural length of the chord

2mgH = 10mg/H x^2
H/√5 = X
H/√5= H-L
L = (`1-1/√5)H
answer for part A

please help me with part b thanks

 

[BvU]

OK, so you have a cord with length $H(1-\sqrt{1/5})$ that has a spring constant $10mg/H$. How much does that stretch when you carefully and slowly hang a mass m from that ?

 

[timetraveller123]

sir i did what you said using differentiation and i got the right answer thanks but why does this work as the method does not involve any friction and yet the question talks about friction what is going on as friction has removed some energy out of the system please explain

 

[haruspex]

sir i did what you said using differentiation and i got the right answer thanks but why does this work as the method does not involve any friction and yet the question talks about friction what is going on as friction has removed some energy out of the system please explain

I would not have termed it friction. The cord will not be perfectly elastic, so some energy is lost in each oscillation. But what we call it is not important here.
If you attach a mass to a spring and lower it carefully, in your hand, say, the weight does work on your hand as well as on the spring.

 

[timetraveller123]

ok thanks sir would you mind helping me with a another problem but it is about special relativity i have been having a lot of problem with special relativity or should i open another thread for that thanks!

 

[haruspex]

ok thanks sir would you mind helping me with a another problem but it is about special relativity i have been having a lot of problem with special relativity or should i open another thread for that thanks!

Definitely a new thread.

 

[timetraveller123]

ok