A 53.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 6.60 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

omega = sqrt (k/m)
so
2 pi f = sqrt (k/m)
so
2 pi/T = sqrt (k/m)
so
2 pi/ 6.60 = sqrt (k/53)
solve for k

then
m g = k x
x = 53*9.81/k