A 51.8-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 (force constant) of the bungee cord?
A) 26.1 N/m
B) 19.6 N/m
C) 16.3 N/m

Question

A  51.8-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 (force constant) of the bungee cord?
A)  26.1 N/m
B)  19.6 N/m
C)  16.3 N/m

Quizplus · Accepted Answer

The period of oscillation $T$ for a mass $m$ attached to a spring with spring constant $k$ in simple harmonic motion is given by the formula $T = 2\pi\sqrt{\frac{m}{k}}$. Rearranging this formula to solve for $k$ gives $k = \frac{4\pi^2m}{T^2}$. Plugging in the given values ($m = 51.8 \, \text{kg}$ and $T = 11.2 \, \text{s}$), we find $k \approx 16.3 \, \text{N/m}$.

A 518-Kg Bungee Jumper Jumps Off a Bridge and Undergoes Simple