Calculate frequency based on change in hanging mass

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Homework Help Overview

The discussion revolves around calculating the frequency of an elastic cord when different masses are hung from it. The original poster presents a scenario where a frequency of 3.0 Hz is observed with a mass of 0.60 kg, and they seek to determine the frequency when a lighter mass of 0.38 kg is used.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between frequency and mass, noting that frequency is proportional to the inverse square root of mass. The original poster attempts to calculate the spring constant and then use it to find the new frequency, while others suggest a more straightforward approach.

Discussion Status

There is an ongoing exploration of different methods to approach the problem. Some participants have provided guidance on simplifying the calculations, while the original poster expresses uncertainty about how to effectively use the initial conditions to find the new frequency.

Contextual Notes

The original poster references a discrepancy between their calculated frequency and the book's answer, indicating a potential misunderstanding or miscalculation in their approach. There is also a mention of the importance of understanding the relationship between mass and frequency without overcomplicating the calculations.

EroAlchemist
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Homework Statement


An elastic cord vibrates with a frequency of 3.0 Hz with a mass of 0.60kg is hung from it. What is its frequency when only 0.38kg hangs from it?


Homework Equations


freq = (1/2Pi)*sqrt(k/m)

The Attempt at a Solution



freq = 3.0Hz = (1/2Pi)*sqrt(k/0.60kg)
= (4Pi^2)(9Hz)(.60kg) = k = 853.7 Nm

use k to calc freq:
freq = (1/2Pi)*sqrt(853.7Nm/0.38kg)
Freq = 7.54 Hz

Book gives correct answer as being 3.8 Hz. This seemed simple enough, but I'm obviously missing something. Thanks much for any help.
 
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Hi EroAlchemist! :wink:

(have a square-root: √ and a pi: π )

the only thing that matters is that it's proportional to 1/√m

nothing else matters, there's no need to do all those calculations to find k and then to eliminate it again …

it just takes time and risks making a mistake!

start again :smile:
 
Thanks tiny-tim!
I used f = (1/2π)(√1/.22kg) = 3.35 Hz
That's a closer answer to the book's 3.8Hz answer. However, by not figuring k using the information that the .60kg weight made the spring oscillate at a frequency of 3Hz, I'm not sure how to use the information about the .60kg weight and the 3Hz frequency.

Thanks for the √ and the π
:)
EA
 
it's proportional to 1/√m …

start with 3 Hz, then adjust for the different m
 

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