Question

A 75 kg student jumps off a bridge with a 12-m-long bungee cord tied to his feet. The massless bungee cord has a spring constant of 430 N/m. a. How far below the bridge is the student’s lowest point? b. How far below the bridge is the student's resting position after the oscillations have been fully damped?

Answer 1

Let us take the gravitation reference point in this question on the bridge.

So, the initial energy of the system is

Now, let us assume the lowest point is a distance l below the bridge. So, the energy of the system at the lowest position can be given by

The conservation of energy gives us

The solution for the above quadratic equation (l > 12) is

So, the lowest point is 20.338 m below the bridge.

Now,

At equilibrium, the forces on the student are zero. i.e

So, the equilibrium point of the student is 13.709 m below the bridge.