# Oscillation problem

1. Apr 26, 2005

### 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 lenght 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?

2. Apr 26, 2005

### Staff: Mentor

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

3. Apr 26, 2005

### OlderDan

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?

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