Finding frequency if you know mass and length of rope.

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SUMMARY

The discussion focuses on calculating the frequency of a girl swinging from a rope with a mass of 40 kg and a length of 2.8 m. The correct approach involves using the formula for a simple pendulum, T = 2π√(L/g), where L is the length of the rope and g is the acceleration due to gravity (9.8 m/s²). The calculated period is 3.35 seconds, leading to a frequency of approximately 0.299 Hz. This method effectively demonstrates the relationship between period and frequency in pendulum motion.

PREREQUISITES
  • Understanding of simple pendulum motion
  • Familiarity with the formula T = 2π√(L/g)
  • Basic knowledge of frequency and period
  • Concept of gravitational acceleration (g = 9.8 m/s²)
NEXT STEPS
  • Explore the derivation of the simple pendulum formula
  • Learn about the differences between simple and physical pendulums
  • Investigate the effects of mass on pendulum motion
  • Study harmonic motion and its applications in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and pendulum dynamics, as well as educators looking for practical examples of oscillatory motion.

astru025
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Homework Statement



A girl with a mass of 40 kg is swinging from a rope with a length of 2.8 m. What is the frequency of her swinging?

Homework Equations



I could find the period of the situation and then from there could calculate the frequency. For the period I'm not sure what equation I should use. I tried using the equation T=2∏ (√I / mgh ). This did not work for me though. Any help would be nice!

The Attempt at a Solution


Attempt shown above.
 
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astru025 said:
I tried using the equation T=2∏ (√I / mgh ). This did not work for me though.
That's the formula for a physical pendulum, which should work just fine. Show what you did.

Of course, since you are ignoring the mass of the rope (I presume), you can also treat this as a simple pendulum.
 
Okay I used the equation for a simple pendulum: T= 2 x pie x square root of L / g. Square root of 2.8 / 9.8 x 2pie gave me 3.35 for a period. Then I plugged in 1/3.35 for the frequency and came up with the right answer. Thanks so much!
 

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