91Ó°ÊÓ

The spring stiffness constant value can be obtained by using the period equation of Simple Harmonic Motion for a mass on a spring. In the time period formula, the value of period of oscillation of the bungee jumper can be obtained by taking the ratio of the required time to reach a low point and the number of oscillating times.

Given data:

The mass of the bungee jumper is \(m = 65.0\;{\rm{kg}}\).

The time of motion is \(t = 43.0\;{\rm{s}}\).

The distance to come to rest below the level of the bridge is \({l_2} = 25.0\;{\rm{m}}\).

The period of oscillation of the bungee jumper can be calculated as:

\(\begin{aligned}{c}T &= \frac{t}{n}\\T &= \frac{{43.0\;{\rm{s}}}}{7}\\T &= 6.14\;{\rm{s}}\end{aligned}\)

The spring stiffness constant can be calculated as:

\(\begin{aligned}{c}T &= 2\pi \sqrt {\frac{m}{k}} \\\left( {6.14\;{\rm{s}}} \right) &= 2\pi \sqrt {\frac{{65.0\;{\rm{kg}}}}{k}} \\{\left( {\frac{{6.14\;{\rm{s}}}}{{2\pi }}} \right)^2} &= \left( {\frac{{65.0\;{\rm{kg}}}}{k}} \right)\\k &= 68.13\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}\end{aligned}\)

The extension of the cord can be calculated as:

\(\begin{aligned}{c}F &= k\Delta x\\mg &= k\Delta x\\\left( {65.0\;{\rm{kg}}} \right) \times \left( {9.81\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}} \right) &= \left( {68.13\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}} \right) \times \Delta x\\\Delta x &= 9.35\;{\rm{m}}\end{aligned}\)

The unstretched length of the bungee cord can be calculated as:

\(\begin{aligned}{c}\Delta x &= {l_2} - {l_1}\\\left( {9.35\;{\rm{m}}} \right) &= \left( {25\;{\rm{m}}} \right) - {l_1}\\{l_1} &= 15.65\;{\rm{m}}\end{aligned}\)

Thus, the spring stiffness constant and the unstretched length of the bungee cord is \(68.13\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}\) and \(15.65\;{\rm{m}}\) respectively.