Chapter 25, Problem 26P

Three linked bungee jumpers are depicted in Fig. P25.26. If the bungee cords are idealized as linear springs (i.e., governed by Hooke's law), the following differential equations based on force balances can be developed

$\begin{array}{l}{m}_1\frac{d^2{x}_1}{d{t}^2}=m_{1}g+k_2\left(x_2-x_1\right)-k_1{x}_1\\ m_2\frac{d^2{x}_2}{d{t}^2}=m_{2}g+k_3\left(x_3-x_2\right)+k_2\left(x_1-x_2\right)\\ m_3\frac{d^2{x}_3}{d{t}^2}=m_{3}g+k_3\left(x_2-x_3\right)\end{array}$

where $m_i=$ the mass of jumper i (kg), $k_j=$ the spring constant for cord j (N/m), $x_i=$ xi 5 the displacement of jumper i measured downward

from its equilibrium position (m), and $g=$ gravitational acceleration (9.81 m/s²). Solve these equations for the positions and velocities of the three jumpers given the initial conditions that all positions and velocities are zero at $t=0$. Use the following parameters for your calculations: $m_1=60\text{ kg, }{m}_2=70\text{ kg,}{m}_3=80\text{ kg, }{k}_1=k_3=50\text{ and }{k}_2=100\left(\text{N/m}\right)$