Abdul Azis P.

asked • 06/22/21# Oscillating Motions

A 65 kg person wants to try bungee jumping. He went to an adventure park with a bungee jumping activity. Then, he jumped from a very high bridge where the bungee rope is tied, reaching the lowest point 8 times through the process of oscillation that spends 38 seconds. The terrifying activity comes to a stop while the person is hanging 25 meters below the level of the bridge. Calculate the spring stiffness constant and the unstretched length of the bungee rope. Assume that the oscillation follows simple harmonic motion.

## 1 Expert Answer

Anthony T. answered • 06/22/21

Patient Math & Science Tutor

There is a change in potential energy of mgΔh from the the bridge to the lowest point = 65kg x 9.8ms^{-2} x 25m

= 15925 j.

At the lowest point, this change in gravitational PE is converted to elastic PE = 1/2 k D^{2} where D is the amount the rope is stretched. So, we can write 65kg x 9.8ms^{-2} x 25m = 1/2 kD^{2}. There are two unknowns in that equation, k the stiffness constant and D the stretch.

To find k we can use T = 2pi x √m/k where T is the period of oscillation. T = 38s/8 = 4.75 s. Now we can calculate k = 4 pi^{2}m / T^{2} = 113.6 N/m. Now we can solve the above equation for D

65kg x 9.8ms^{-2} x 25m = 1/2 kD^{2} = 15925 = 1/2 x 113.6 x D^{2} or D = 16.7 m.

Therefore, the unstretched rope is 25m -16.7m = 8.3 m.

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Anthony T.

Are we to assume that there is no slack in the rope and that it begins to stretch as soon as the person leaves the bridge?06/22/21