A 55.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 7.20 s. The elastic rope obeys Hooke's Law. By how much is the rope extended beyond its unloaded length when the performer hangs at rest?

Answer 1

The period of oscillations is
T = 2•π•sqr(m/k)
k= 4• π^2•m/T^2 =41.84 N/m

Hook’s law: F = - kx, F=-mg
mg = kx,
x=mg/k=55•9.8/41.84 =12.88 m

Answer 2

Thank you , Elena.

Answer 3

Thank you!