Bungee Jumper A 65.0 kg bungee jumper jumps off a high bridge, attached to a bun

Question

Bungee Jumper A 65.0 kg bungee jumper jumps off a high bridge, attached to a bungee cord whose unstretched length is 11.0 m. The other end of the bungee cord is attached to the bridge. The bungee jumper reaches the bottom of her motion 36.0 m below the bridge before bouncing back up. Her motion can be separated into an 11.0 m free fall and a 25.0 m section of simple harmonic motion. The goal of this problem is to find the time between when she jumps off and the bottom point. What is the time interval for free fall (the first 11.0 m)? Use energy conservation to find the spring constant of the bungee cord. How far below the bridge is the point where the bungee cord has engaged and exerting an upward force that equals the weight of the jumper - i.e., where there is zero net force on the jumper? This will be the "origin" point in your description of simple harmonic oscillation. What is the angular frequency of oscillation? [Remember, as a constant force gravity will not affect the frequency of oscillation - it only shifts the zero-force point downward according to the amount you found in part (c).] Find the time interval for the cord to stretch 25.0 m, and then the total time of the fall. Useful it will be, for this part, to make a sketch of the jumper's vertical position as a function of time. Use the equilibrium position of part (c) as y = 0. Label clearly the regions of free fall, and of simple harmonic motion, and then figure out the amplitude of oscillation for the SHM part, and identify the time interval you are looking for.

Explanation / Answer

a) time for first 11m

S = ut + 0.5at2

11 = 0*t + 0.5*9.8*t2

so t=1.49sec

b) total energy before jumping = P.E. at 36m height (taking lowest pt. as reference)

so initial energy = mgh = 65*9.8*36 = 22932J

this energy is stored as elastic P.E. in string = 0.5kx2

so k= 22932/0.5*252 = 73.38 N/m

c) let length at which force applied by string is equal to the weight = L

so kL = mg

73.38*L = 65*9.8

L= 8.68m So length below bridge = 11+8.68 = 19.68m

d) angular frequency = k/m = 73.38/65 = 1.11289 sec-1

e) time period = 1/f = 1/0.169 = 5.91 sec

this time is for one up and down motion so for one way time = 5.91/2 = 2.95 secs

So total time = 2.95+1.49 = 4.44 secs

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