How do you find the length of a rope

A 55.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 4.00 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?

can someone show me how to do this?

Using Hookes law we find the spring constant k
from formula for natural frequency

fn=1/2pi sqrt(k/m)
once every four seconds is 0.25 cycles per second so

0.25 = 1/2pi sqrt(k/55) solve for k
(0.25(2pi))^2(55) = k
k = 135.7 N/m

so for hanging from rest on the spring constant k
F = kx
55(9.81) = 135.7x
x = 4 m below unstressed length

Idk