A bungee jumper, whose mass is 87 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 8.5 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 need to use the equation for the period of oscillation of a mass-spring system.
The period (T) is the time taken for one complete oscillation. In this case, we can see that the jumper reaches the lowest point three times in a total of 8.5 seconds. Therefore, the period can be calculated by dividing the total time (T_total) by the number of oscillations (n).
T = T_total / n
T = 8.5 s / 3
T ≈ 2.83 s
Next, we can use the equation for the period of a mass-spring system:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant. Rearranging the equation to solve for k:
k = (2π)^2 * (m / T^2)
k = (2π)^2 * (87 kg / (2.83 s)^2)
k ≈ 1047 N/m
Therefore, the spring constant of the bungee cord is approximately 1047 N/m.
If he reaches the low point twice in 8,5 s, the period of oscillation is 4.25 s.
Use the formula for the Period to compute the value of the spring constant k
P = 2 pi sqrt (m/k)