A 56.0 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?

Question

Quizplus · Accepted Answer

The period of oscillation for a mass-spring system is given by:$$T = 2\pi \sqrt{\frac{m}{k}}$$where:* T is the period of oscillation in seconds* m is the mass in kilograms* k is the spring constant in N/mSolving for k, we get:$$k = \frac{4\pi^2 m}{T^2}$$Plugging in the given values, we get:$$k = \frac{4\pi^2 (56.0 	ext{ kg})}{(11.2 	ext{ s})^2} = 17.6 	ext{ N/m}$$

A 560 Kg Bungee Jumper Jumps Off a Bridge and Undergoes