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Complex Physics Problem
Name: Pavan
Status: student
Age: 17
Location: N/A
Country: N/A
Date: 1999
Question:
Hello, this is a problem in my physics book, and I couldn't
figure it out. It has to do with SHO and SHM:
A bungee jumper with mass of 65.0kg jumps from a high bridge. After
reaching his lowest point, he oscillates up and down, hitting a low point
eight more times in 34.7 s. He comes to rest 25.0 m below the level of the
bridge. Calculate the spring constant and the unstretched length of the
bungee cord.
These are the answers, I just don't know how to reach them.
k= 136 N/m L= 20.3m
Replies:
Regarding the problem posed by the student above, the solution is as
follows. First, let us note that the fact that motion is damped
harmonic has no relevance to the problem, so air resistance is not a
consideration.
The period of SHM is just 34.7/8, and this has to be equated to the
formula for the period of an oscillating spring, 2*pi*sqrt{m/k}
where m is the mass of the person, and k is the force constant of
the spring. Solving for k gives the answer the student claims.
For unstretched length, the equilibrium condition when the bungee
has ceased oscillating in the vertical direction is that the
person's weight, mg, must equal the bungee's elastic force. Now
elastic force = kx, where x is the stretch on the cord. Equating, mg
= kx gives x, which you then subtract from 25 meters to get the
unstretched length.
Regards,
dr. viv
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