# A bungee jumper with mass 58.5

A bungee jumper with mass 58.5 kg jumps from a high bridge.After arriving at his lowest point, he oscillates up and down,reaching a low point seven more times in 41.0 s . He finally comesto rest 20.0 m below the level of the bridge. Part A Estimate thespring stiffness constant of the bungee cord assuming SHM. Part BEstimate the unstretched length of the bungee cord assumingSHM.

*The bungee jumper has a mass (m) of 58.5 kg*He hits a low point eight times after his initial in 41seconds.*His resting point, when he stops oscillating, is 20 meters belowthe level of the bridge where he began, so this is the equilibriumpoint of the spring with the jumper attached to it.For the first part A,

we need to find the spring constant of the bungee cord, and thisis easily found through the equation:T = (2π)*sqrt(m/k)Where T is the period of the oscillation, m is the mass of thespring, and k is the spring constant.So, we have the mass of the spring, which is just 58.5 kg, becausethe mass of the bungee cord itself is negligible in this situation.But we need to find the period T before we can solve for k.The period of the spring is how long it takes an object to go frompeak to peak in an oscillation. And since we know that the jumperpeaked 7 times in 41 seconds, we can simply divide 41 s by 7 tofigure out about how long it takes the jumper to go from low-pointto low-point. So:(41 s)/(7) = 5.857 s = TNow we can take all of out information and solve for k:T = (2π)*sqrt(m/k)(5.857 s) = (2π)*sqrt((58.5 kg)/k)Dividing both sides by (2π) and then squaring both sidesgives:((5.857 s)/(2π))^2 = (sqrt((58.5 kg)/k))^2(0.8698 s^2/m^2) = (58.5 kg)/kk = 67.255 N/m = spring constant of bungee cordPart B)

Now that we have our spring constant, we can easily find theunstretched length of the bungee cord.To do this, the stretch of the cord can be solved by setting theforce of the jumper pulling down on the cord equal to the force ofthe spring pulling up on the jumper. So:mg = kΔdWhere m is the mass of the jumper, g is the gravitationalacceleration, k is the spring constant, and Δd is the change indistance of the bungee cord (with vs without the jumper attached toit)Plugging in what we know:(58.5 kg)(9.81 m/s^2) = (67.255 N/m)(Δd)Simplifying and dividing both side by (67.255 N/m) gives:8.533 m = ΔdSince this is the change in equilibrium point with vs. without thejumper, we can simply subtract this value from 20 meters, which isthe equilibrium point the cord experiences when the jumper is stillattached to it. So:(20 m) – (8.533 m) = 11.467 mThis is the length of the unstretched bungee cord.

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