Simple Harmonic Motion - At which rate will a ball vibrate

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SUMMARY

The discussion focuses on calculating the frequency of vibration for a 60.0 kg ball suspended from a 1.80 m long wire that stretches 2.00 mm under load. The spring constant was determined to be 294300 N/m using the formula F=kx. The period of oscillation was calculated as 0.0897 seconds, leading to a frequency of 11.15 Hz, confirming the accuracy of the calculations. The stress on the wire remains below the proportional limit, validating the assumptions made in the calculations.

PREREQUISITES
  • Understanding of Hooke's Law (F=kx)
  • Knowledge of the formula for the period of oscillation (T=2π√(m/k))
  • Ability to calculate frequency from period (f=1/T)
  • Familiarity with concepts of stress and proportional limits in materials
NEXT STEPS
  • Study the derivation of the spring constant in different materials
  • Explore the effects of varying mass on the frequency of oscillation
  • Investigate the relationship between wire length and frequency in simple harmonic motion
  • Learn about damping effects in oscillatory systems
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Students in physics, mechanical engineers, and anyone interested in the principles of simple harmonic motion and oscillatory systems.

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A hanging wire is 1.80 m long. When a 60.0 kg ball is suspended from the wire, the wire stretches 2.00 mm. If the ball is pulled down a small additional distance and released, at what frequency will it vibrate? Assume that the stress on the wire is less than the proportional limit.
I have found an answer but it seems too simple and does not take into account the length of the wire but I am pretty sure that it should. any help is appreciated. thanks


Homework Equations



F=kx
T=2pi*sqrt(m/k)
f=1/T

The Attempt at a Solution


Found the spring constant to be 294300n/m

then found the period of oscillation to be 0.0897s
then inverted it to find the frequency to be 11.15hz
 
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Yes that is correct.
 

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