Oscillation of rope problem

In summary, a circus performer weighing 55.0 kg oscillates at a rate of once every 8.40 s on a long elastic rope obeying Hooke's Law. Using the equation t=2\pi\sqrt{m/k}, the performer's k value was found to be 0.316. However, this may not be a reasonable estimate as 8.4/6 is approximately 1.4, indicating that k should be larger than m.
  • #1
Punchlinegirl
224
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 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?
 
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  • #2
Punchlinegirl said:
Using the equation [tex] t= 2\pi\sqrt{m/k} [/tex], I solved for k and got .316.
That equation is OK, but your answer for k is not. Redo it.
 
  • #3
Punchlinegirl said:
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?

Do an estimate to see if your k value is reasonable

[tex] T/2\pi = \sqrt{m/k} [/tex]

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?
 

Related to Oscillation of rope problem

1. What is the oscillation of rope problem?

The oscillation of rope problem refers to the phenomenon of a rope or string being attached to two fixed points and then being pulled and released to create a swinging motion. This motion is known as oscillation and is caused by the tension and elasticity of the rope.

2. What factors affect the oscillation of a rope?

The oscillation of a rope can be affected by several factors, including the length and thickness of the rope, the distance between the fixed points, the initial angle of release, and the force applied to the rope. These factors can impact the frequency, amplitude, and period of the oscillation.

3. How is the period of a rope's oscillation calculated?

The period of a rope's oscillation can be calculated using the formula T = 2π√(L/g), where T is the period, L is the length of the rope, and g is the acceleration due to gravity. This formula assumes ideal conditions and a small angle of release.

4. How does the amplitude of the oscillation change over time?

The amplitude of the oscillation decreases over time due to the dissipation of energy through friction and air resistance. This means that the rope's swings will become smaller and smaller until it eventually comes to rest.

5. How can the oscillation of a rope be applied in real-life situations?

The oscillation of a rope can be applied in various real-life situations, such as in pendulum clocks, musical instruments like guitars and violins, and amusement park rides. It is also used in engineering and construction for the design of suspension bridges and tall buildings to ensure their stability during wind and seismic activities.

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