A 56 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion.If the period of oscillation is 11.2 s,what is the spring constant of the bungee cord,assuming it has negligible mass compared to that of the jumper?

Question

Quizplus · Accepted Answer

The spring constant is given by k = (4π²m)/T², where m is the mass of the jumper and T is the period of oscillation. Plugging in the given values, we get k = (4π²(56 kg))/(11.2 s)² = 17.6 N/m.

A 56 Kg Bungee Jumper Jumps Off a Bridge and Undergoes