Simple Harmonic Motion of a Bungee Jumper
Any help with this problem would be great!
A bungee jumper, whose mass is 82 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 9.6s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.
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In case you do not yet know the customary notations:
a^b means "a to the power b",
a_b means " a with subscript b",
sqrt(a) means "square root of a".
The time passing between two consecutive low points is the period of oscillations.
As the 9.6 s were needed to reach the low point on the 3rd time, we deduce that 9.6 ...
Solution Summary
This solution contains step-by-step calculations to determine the spring constant of the bungee cord using angular frequency and harmonic motion.