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Circus Performer on a Rope

  • Thread starter joeypeter
  • Start date
5
0
A 55.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 2.60 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 help me to find the answer step by step?
 

Answers and Replies

23
0
The period of the system (since it obeys Hooke's law) is given by T = 2*pi*sqrt(m/k). You can use the given data to solve for k, the effective spring constant.

When the performer is at rest, the net force on him must be zero. Therefore, the upward spring force, kx, must equal the gravitational force on the person.
 

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