A bungee jumper, whose mass is 83 kg, jumps from a tall platform. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in 9.7 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.
To determine the spring constant of the bungee cord, we can use the formula for the period of an oscillating mass-spring system:
T = 2π√(m/k)
where:
T is the period of the oscillation
m is the mass of the bungee jumper
k is the spring constant
In this case, we are given the period T and the mass m, and we need to solve for the spring constant k.
First, let's calculate the period using the information given. We are told that the bungee jumper reaches the lowest point two more times in 9.7 seconds, which means the time it takes for one complete oscillation (T) is 9.7 seconds divided by 3 (since there are three low points: the initial lowest point, and two more low points).
T = 9.7 s / 3 ≈ 3.233 s
Now we can rearrange the formula and solve for the spring constant k:
T = 2π√(m/k)
Squaring both sides:
T^2 = (4π^2)(m/k)
Rearranging further:
k = (4π^2m) / T^2
Plugging in the given values:
k = (4π^2)(83 kg) / (3.233 s)^2
Evaluating the expression:
k ≈ 1015 N/m
Therefore, the spring constant of the bungee cord is approximately 1015 N/m.