Question

A 90 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 420 N/m. You can assume that the bungee cord exerts no force until it begins to stretch.

How far below the bridge is the student's lowest point?

How far below the bridge is the student's resting position after the oscillations have been fully damped?

Answer 1

As he falls 12 m , cord comes to original length. But since the student has kinetic energy so he falls further down till kinetic energy becomes zero.

Suppose it further falls by y and then stops momentarily. Let the Cord is stretched by y metre at this moment.

By energy conservation,

Decrease in gravitational potential energy = gain in elastic potential energy

mg (12 + y) = (1/2) k y²

90 × 9.8 × ( 12 + y) = (1/2) × 420 × y²

10584 + 882 y = 210 y²

Therefore

210 y² - 882 y - 10584 = 0

Solving we get

y = 9.5 m

Therefore lowest point is 12 + 9.5 = 21.5 m below the bridge

Finally oscillation will stop due to damping and student will be in equilibrium position and cord will be stretched to balance the weight.

Suppose stretching is x

Force applied by cord = k x, in upward direction

Therefore

k x = mg

420 x = 90 ×9.8

Therefore

x = 2.1 m

So in equilibrium position

Student will be 12 + 2.1 = 14.1 m below the bridge.