Bungee Jumping: Solving for Mass, Cord Length, and Period of Oscillations

AI Thread Summary
The discussion revolves around calculating the period of oscillations for a bungee jumper using Hooke's Law, with a focus on a 58.0 kg jumper and a 20.0 m bungee cord. The calculated period of oscillation is approximately 13.6 seconds. Concerns arise regarding the suitability of the same cord for an 88.0 kg jumper, as the increased mass would cause the cord to stretch more, potentially leading to a dangerous situation. The correct interpretation of the stretch distance is emphasized, clarifying that the initial cord length is 20 meters, not 46 meters. The conclusion is that the heavier jumper should not use the same cord due to the risk of hitting the water.
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Homework Statement



A bungee jumper leaps from a bridge and undergoes a series of oscillations. Assume g=9.78 m/s^2. If a 58.0 kg bungee jumpers jumps from length of 20.0 m and she jumps from a heigh of 66.0 m above the river, coming to rest a few centimeters above the water surface on the first downward descent. What is the period of the oscillations? Assume the bungee cord follows Hooke's Law. The next jumper in line has a mass of 88.0 kg. Should he jump using the same cord?

A) kx - mg= 0

k=mg/x
k= (58.0 * 9.78)/46
k=12.33
w= sq rt (k/m)=sq rt (12.33/58.0)=.4611
w=2pif
w=2pi/T
T = (2pi)/w
T=(2pi)/.4616
T=13.6 seconds

B) kx -mg =0
kx=mg
x=mg/k
(88.0 * 9.78)/12.36=69.6

No, he should not jump using the same cord b/c the cord will stretch and he will not be able to avoid the water.

These answers are wrong and I don't see why.

Homework Equations





The Attempt at a Solution

 
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x is not 46 metres. It is 20 metres because the x stands for how much it stretched. If that doesn't work, try 1/2kx^2.

46 metre is the normal length of the rope.
 
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