Phys with Calc 1. A 75.0 kg bungee jumper steps off a bridge with a light bungee

Question

Phys with Calc 1.

A 75.0 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) in freefall, distance = 1/2 gt^2
L=distance=12m
t = sqrt (2 distance / g)
=sqrt(2*12/9.8)

=1.56 s

b) initial gravitational PE = mgx
= final spring PE = 1/2 k(x-L)^2
x=37 m
k = 2mgx/(x-L)^2

=2*75*9.8* 12/(37-12)^2

=78.4 N/m

c) Here we just use Hooke’s law and balance the vertical forces on the jumper. mg = kLeq

where Leq is the distance from bungee-engage to equilibrium.

Leq=mg/k=75*9.8/78.4=9.375

The equilibrium point is thus, L + Leq = 12+9.375 m = 21.375 m

d) angular frequency is:
omega = sqrt (k/m) =sqrt (78.4/75)=1.02 rad/sec

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