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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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