How Much is the Rope Extended When the Circus Performer Hangs at Rest?

[Punchlinegirl]

A 55.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 8.40 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?

I drew a free body diagram and summed up the forced to get:
$$F_{t} - mg= 0$$
Since $$F_{t}= -kx$$ , I substituted it in.
Using the equation $$t= 2\pi\sqrt{m/k}$$, I solved for k and got .316. I plugged this into the equation and got x=1705 m, which isn't right... can someone tell me what I'm doing wrong?

 

[Doc Al]

Using the equation $$t= 2\pi\sqrt{m/k}$$, I solved for k and got .316.

That equation is OK, but your answer for k is not. Redo it.

 

[OlderDan]

A 55.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 8.40 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?

I drew a free body diagram and summed up the forced to get:
$$F_{t} - mg= 0$$
Since $$F_{t}= -kx$$ , I substituted it in.
Using the equation $$t= 2\pi\sqrt{m/k}$$, I solved for k and got .316. I plugged this into the equation and got x=1705 m, which isn't right... can someone tell me what I'm doing wrong?

Do an estimate to see if your k value is reasonable

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

must be about 8.4/6 which is about 1.4

How must the numerical value of k compare to the numerical value of m?