Question

A bungee jumper with mass 60.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 49.0 s. He finally comes to rest 30.0 m below the level of the bridge.
1) Calculate the spring constant of the bungee cord.2) Calculate the unstretched length of the bungee cord.

Answer 1

Answer:

a) Spring constant,k = 63.09 N/m

b) Unstretched length = 20.68 m          

Explanation:

a) Mass of the bungee jumper, M = 60 kg

time for complete oscillation, t = 49 s

Number of oscillations, n = 8

The period of oscillation, T = t/n

T = 49/8

T = 6.125 s

To calculate the spring constant of the bungee cord, use the formula:

$T = 2\pi \sqrt{\frac{M}{k} }$

$6.125 = 2\pi \sqrt{\frac{60}{k} } \\6.125/2\pi = \sqrt{\frac{60}{k} }\\0.975^2 = 60/k\\0.951 = 60/k\\k = 60/0.951$

Spring constant,k = 63.09 N/m

b) The extension in the spring is given by the formula:

x = mg / K

x = (60*9.8)/63.09

x = 9.32 m

extension, x = final length - original length

Original length = Final length - extension

Original length = 30 - 9.32

Original length = 20.68m                                

Unstretched length = 20.68 m            

Answer 2

Answer:

1) 10.05 N/m

2) 20.69 m

Explanation:

1) If the jumper completes 8 cycles in 49 seconds, then the period of each cycles is T = 49/8 = 6.125 s

From this info we can calculate the spring constant, treating this as simple harmonic motion k:

$T = 2\pi\sqrt{\frac{m}{k}}$

$\frac{m}{k} = \frac{T^2}{4\pi^2}$

$k = \frac{4\pi^2 m}{T^2} = \frac{4\pi^2 60}{6.125^2} = 63.14 N/m$

2) Let g = 9.8 m/s2. So the jumper weight is F = mg = 60*9.8 = 588 N. With this force the jumper would have stretched the bungee a length of

x = F/k = 588 / 63.14 = 9.31 m

As the jumper is at rest 30m below the bridge and the cord is stretched by 9.31m. Then its original length would be 30 - 9.31 = 20.69 m