A bungee jumper, whose mass is 81 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 8.9 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant
Answer in N/m
To determine the spring constant of the bungee cord, we can use the principles of simple harmonic motion. In this case, the bungee jumper oscillates up and down, behaving like a mass-spring system.
The time period (T) of the oscillation can be calculated using the equation:
T = 2π√(m/k),
where m is the mass of the bungee jumper and k is the spring constant.
In this case, the time period is given as 8.9 seconds. We know the mass (m) of the bungee jumper is 81 kg. We need to calculate the spring constant (k).
Rearranging the equation, we get:
k = (4π²m) / T².
Now, we can substitute the given values and calculate the spring constant.
k = (4π² * 81 kg) / (8.9 s)².
Calculating it further:
k = (4 * 3.1416² * 81 kg) / (8.9 s)².
k ≈ 361.33 N/m.
Therefore, the spring constant of the bungee cord is approximately 361.33 N/m.