- #1
Punchlinegirl
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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:
[tex]F_{t} - mg= 0[/tex]
Since [tex] F_{t}= -kx [/tex] , I substituted it in.
Using the equation [tex] t= 2\pi\sqrt{m/k} [/tex], 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?
I drew a free body diagram and summed up the forced to get:
[tex]F_{t} - mg= 0[/tex]
Since [tex] F_{t}= -kx [/tex] , I substituted it in.
Using the equation [tex] t= 2\pi\sqrt{m/k} [/tex], 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?