bungee jumping and pendulum period.......physics help?

Question

A man jumps off a 20 m high bridge using an elastic bungee cord of unstretched length 10 m. The jump is so designed that the man at the lowest point would just touch the water, and be yanked back up. Find the period, in seconds, of simple harmonic motion that would result.

Find the length in meters of a simple pendulum with the same period.

Answer 1

T=(2 pi)sqrt(m/k)
Potential energy = Energy of the bungee cord

Pe = Eb

mgh = 0.5 k L^2 since h=2L
k=4 m g / L

T=(2 pi)sqrt(m L/(4 m g ))
T=(2 pi)sqrt( L/(4 g ))
We know that L=10 m and g=9.8 m/sec^2
T=(2 pi)sqrt( 10 /(4 x 9.8 ))
T=3.1sec

for pendulum
T=2 pi sqrt(l/g)

l= g (T/(2 pi))^2
l= 9.8 ( 3.1/(2 x pi))^2 =
l=2.4 m

Answer 2

preliminary 2d - he's equipped to bounce off - E and ok are 0. very final 2d G is 0. easily, G could additionally be seen 0 any the area, so it is going to be greater precise to declare that G is minimum at very final 2d and optimum in the commencing up. you're able to be able to desire to argue a "organic 0" for G on the floor of the earth, yet that's variable. The word "somewhat touches water" is, I comprehend, meant to advise "basically approximately the earth's floor". the comparable could be pronounced (mathematically)approximately E, yet there seems to be a greater "organic" 0 for it (unstretched difficulty).