# Circus performer oscillating

1. Jun 9, 2009

### smillphysics

A 57.0 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

T=2pi* sqrt m/k
kx=mg

7.2sec=2pi*sqrt 57/k, then sq. rt k= 2pi*sqrt 57/1.2 , I get k=4.2
then I put k into the second equation to get x=133m which is the wrong answer- What am I doing incorrectly?

2. Jun 9, 2009

### rock.freak667

Period is the number of oscillations in 1 second

so in 7.20 seconds there is 1 oscillation. How many oscillations in 1 second? That will give you T.

3. Jun 10, 2009

### smillphysics

Ok so that gives me 1oscillation in 7.2 sec which is 1/7.2=.139 , then the sq rot of k= 2pi*sqrt m/period which gives me a k=11.28. I plug that into the equation 57*9.8/11.28 and get x=49.5m which is still wrong? So Not sure where I am going wrong here.

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