A 70.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her

Question

A 70.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the cord is 12.0 m. She reaches the bottom of her motion 37.0 m below the bridge before bouncing back. Her motion can be separated into an 12.0 m free-fall and a 25.0 m section of simple harmonic oscillation.

(a) For what time interval is she in free-fall? s

(b) Use the principle of conservation of energy to find the spring constant of the bungee cord. N/m

(c) What is the location of the equilibrium point where the spring force balances the gravitational force acting on the jumper? Note that this point is taken as the origin in our mathematical description of simple harmonic oscillation. m below the bridge

(d) What is the angular frequency of the oscillation? rad/s

(e) What time interval is required for the cord to stretch by 25.0 m? s

(f) What is the total time interval for the entire 37.0 m drop? s

Explanation / Answer

a)"The free-fall phase follows the parabolic

distance = 1/2 gt^2
t = sqrt (2 distance / g)

t=1.56s

b)initial gravitational PE = mgh
= final spring PE = 1/2 kx^2

k = 2mgh/x^2

81.2224 N/m

c) spring force = weight
kx = mg
x = mg / k

x=8.446m

The equilibrium point is thus
distance+ x=12+8.446=20.45 m

d) angular frequency is:
omega = sqrt (k/m)

1.077rad/s

e)the stretch time is 1/4 the total period
T/4 = (2 pi / omega) / 4
= pi / 2omega
=1.458sec

e)Combining the free-fall time from (a) with the harmonic time from (e) we have

1.458+1.56=3.018 secs

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