Setup: Suppose a midshipman (6 feet tall, weighing 160 pounds) intends to jump off of a 206 foot high bridge. He has his choice of bungee cords to tie to his feet – each is 100 feet long, but with varying degrees of stiffness. How stiff a cord should he choose (what k value?) so that he just gets the crown of his head wet in the river below?
The physics: For a first run through, we will ignore air resistance.
The main difference between this problem and the typical spring problems we have tried is the behavior of the cord when the object (or midshipman) it is supporting is above the cord’s natural unweighted length. A spring will, whether an object is above or below equilibrium, push or pull the object toward the equilibrium position. A bungee cord, however, will simply go slack when the object it is holding is above the equilibrium position – it provides no push at all! As such, we get a piecewise restoring force,
where x is oriented downward, as usual, and x = 0 is 100 feet below the bridge (where the bungee cord is at its natural length). See the figure below.
Figure 1: Coordinates for the bungee jumper (care of Zill and Cullen, 6th edition)
If the only forces on the bungee jumper are gravity and the restoring force of the bungee cord, then Newton’s Second Law of Motion can be written as the differential equation
Recall that a person weighing 160 pounds near the earth’s surface has a mass of 5 slugs.
Question 1: How long is the midshipman in freefall? That is, how many seconds pass before the jumper feels any force at all from the bungee cord around his ankles?
Response: In this regime, r(x) = 0, and so the differential equation reduces to about the simplest one can imagine
Assuming the jumper gingerly steps off the bridge at time t = 0, we have initial conditions x(0) = -100 ft and x′(0) = 0 ft/s. One could use annihilators, etc., and the full power of our differential equations course to solve this equation, or one could channel the inner calculus geek and quietly integrate twice to find
x is equal to zero when t = 2.5 seconds.
Question 2: How stiff should the cord be so that the jumper just dunks the crown of his head into the water?
Response: After t = 2.5, the jumper feels the restoring force of the bungee cord, and the differential equation becomes
where k depends upon the particular bungee cord we choose. Note that our initial conditions here come from the previous calculations: x(2.5) = 0 and x′(2.5) = 80 ft/s. This is a second order, nonhomogeneous equation, and now we can unapologetically find a solution using the method of undetermined coefficients (annihilators):
Since our jumper is 6 feet tall, and the bridge is 206 feet tall, he will just dampen the top of his head when the bungee cord stretches to exactly 100 feet beyond its natural length. In this case, we want the amplitude of the oscillation to be 100 – 160/k. So we can set
Squaring both sides, the k2-terms cancel and we are left with a linear equation in k. Solving, we find that k = 6.4 pounds/foot.
Note that with no damping force, the bungee jumper continues to oscillate, alternately moistening his brow and then crashing back into the bottom of the bridge, indefinitely. He follows a traditional parabolic path when x is less than 0 (and there is slack in the cord), and a sinusoidal path when x is greater than 0 (and the cord is taut).
Figure 2: The motion of the undamped bungee jumper.
Extra Credit: Suppose the bungee jumper knows that the force of air resistance in his case is numerically equal to his instantaneous velocity. Including the damping force of air resistance allows for the use of a less stiff (lower k) bungee cord. (See Figure 3, below.) Extend the analysis above to determine what k-value cord the jumper can use to once again just dampen the crown of his head?
Figure 3: The motion of the damped bungee jumper.