A bungee jumper with mass 69.5 kg jumps from a high bridge.
After reaching his lowest point, he oscillates up and down,
hitting a low point eight more times in 36.6 s. He finally
comes to rest 24.3 m below the level of the bridge. Calculate
the spring constant of the bungee cord.
The frequency is 8/36.6 = 0.2185 Hz
That equals [1/(2 pi)]sqrt(k/m)
You know m = 69.5 kg. Solve for k
The 24.3 m number you do not need.
Thanks for the timely help
To calculate the spring constant of the bungee cord, we can use the formula for the period of a simple harmonic motion:
T = 2π√(m/k)
Where:
T is the period of oscillation,
m is the mass of the bungee jumper, and
k is the spring constant.
We are given that the bungee jumper hits a low point eight more times in 36.6 seconds. Since the period (T) is the time taken for one complete cycle, we can calculate it by dividing the total time (36.6 seconds) by the number of cycles (8 + 1, because the initial jump also counts as a cycle):
T = 36.6 s / (8 + 1) = 36.6 s / 9 ≈ 4.07 s
Now, we can rearrange the formula for the period to solve for the spring constant (k):
k = (4π²m) / T²
Substituting the given values:
m = 69.5 kg
T = 4.07 s
k = (4π² * 69.5 kg) / (4.07 s)²
Using the value of π (pi) as approximately 3.14, we can plug in the values and calculate:
k ≈ (4 * 3.14² * 69.5 kg) / (4.07 s)²
k ≈ (4 * 9.86 * 69.5 kg) / (4.07)²
k ≈ (38.52 * 69.5 kg) / (16.5649)
k ≈ 2675.14 kg m/s²
Therefore, the spring constant of the bungee cord is approximately 2675.14 kg m/s².