Harmonic Motion - Bungee Jumping

[youngling]

Homework Statement

You (mass m) are chosen as the "test pilot" for a new bungee jump. When
you suspend yourself at rest from the bungee, you find that it stretches a
distance L under your body weight. The jump point is a height 6L above the
ground and the relaxed (unstretched length) of the bungee is 3L. Should you
make the test jump?

Variables: k - spring constant
L - distance from jump point
I think that's it...

Given: m - your mass
equilibrium length of bungee+mass: 3L + L = 4L

Homework Equations

F = -kx
F = mg
x = A cos(wt + phase)

The Attempt at a Solution

I have no idea but I know that we're dealing with a harmonic motion problem where the bungee cord acts as a spring, and I think we need to determine the amplitude so that we can see if the "lowest" endpoint of the oscillation is greater than 6L from the jump point... because if it isn't then SPLAT. However, I don't know where to start. I said that F = -kx - mg, and then set F = 0 and x = 4L for the equilibrium point and got that k = -mg/4L

 

[andrevdh]

You should rather think in energy terms - energy is stored in the cord when it is stretched. Check what extension is required to absorb all of the potential energy of the test pilot.

 

[m2003]

. But I'm not sure if that's right or how it helps me find the amplitude.I would approach this problem by first defining the variables and equations involved. We have the mass of the person (m), the spring constant (k), the distance from the jump point (L), and the height of the jump (6L). The equations we can use are F = -kx, F = mg, and x = A cos(wt + phase), where A is the amplitude of the oscillation and w is the angular frequency.

Next, I would analyze the forces acting on the person at different points of the jump. At the start of the jump, the person is at rest and the only force acting on them is their weight, mg. As they fall, the bungee cord begins to stretch and exerts a force on the person, in the opposite direction to their motion. This force can be calculated using F = -kx, where x is the distance the cord has stretched. At the bottom of the jump, when the person reaches the lowest point of their oscillation, the bungee cord is fully stretched and exerts a force equal to mg, in the upwards direction.

Now, we can use the equations to find the amplitude of the oscillation. At the equilibrium point, where the person is at rest, we have F = -kx - mg = 0. This gives us the equation kx = mg, which we can substitute into the equation for the amplitude, x = A cos(wt + phase). This gives us A = mg/k. Substituting in the value we found for k in terms of m and L, we get A = 4mg/3L.

Finally, we can use this value for the amplitude to determine if the person will hit the ground during the jump. Since the jump point is 6L above the ground, the distance from the jump point to the lowest point of the oscillation is 3L. If the amplitude of the oscillation is less than this distance, the person will safely complete the jump. However, if the amplitude is greater than 3L, the person will hit the ground and the jump should not be attempted.

In conclusion, based on the given information and using the principles of harmonic motion, we can determine that the person can safely make the test jump if the amplitude of the oscillation is less than 3L.