How Much Does the Rope Stretch When a Circus Performer Hangs at Rest?

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To determine how much the elastic rope stretches when the circus performer hangs at rest, first calculate the spring constant (k) using the formula for the period T = 2*pi*sqrt(m/k), where m is the mass of the performer and T is the oscillation period. Rearranging this formula allows for solving k with the provided values. When the performer is at rest, the upward spring force (kx) must balance the downward gravitational force (mg), leading to the equation kx = mg. Solving for x gives the extension of the rope beyond its unloaded length. This analysis illustrates the relationship between mass, gravitational force, and the properties of the elastic rope in a circus setting.
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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 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?
 
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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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