Simple Harmonic Motion - At which rate will a ball vibrate

AI Thread Summary
The discussion focuses on calculating the frequency of a ball vibrating from a hanging wire that stretches under its weight. A 60.0 kg ball causes a 2.00 mm stretch in a 1.80 m long wire, leading to the determination of the spring constant as 294300 N/m. The period of oscillation is calculated to be 0.0897 seconds, resulting in a frequency of 11.15 Hz. The initial concern about the simplicity of the answer is addressed by confirming the calculations align with the principles of simple harmonic motion. The solution effectively demonstrates the relationship between mass, spring constant, and frequency in this context.
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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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