How to find fundamental frequency

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Homework Help Overview

The problem involves determining the new fundamental frequency of a sculpture submerged in water, initially suspended from a steel wire. The context includes concepts from mechanics and wave physics, particularly focusing on tension in the wire and its effect on wave frequency.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Archimedes' principle to find the new tension in the wire after submerging the sculpture. There is a focus on the relationship between tension and frequency, with some questioning the relevance of the sculpture's mass in the calculations.

Discussion Status

Some participants have provided guidance on how to relate the old and new frequencies through tension ratios. Multiple interpretations regarding the significance of the sculpture's mass are being explored, with no explicit consensus reached.

Contextual Notes

Participants note the density of water and aluminum as relevant factors in the calculations, and there is an emphasis on the assumptions made regarding the tension in the wire before and after submersion.

ngcg
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Homework Statement


When a 70kg aluminum (density = 2.7 g/cm3) sculpture is hung from a steel wire, the fundamental frequency for transverse standing waves on the wire is 250.0 Hz. The sculpture (but not the wire) is then completely submerged in water. What is the new fundamental frequency?


Homework Equations


f=ma
fundamental frequency = (1/2L)((Tensional force/linear density)^1/2)


The Attempt at a Solution


250=(1/2L)((686N/u)^1/2))
(stuck there)
 
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According to the Archimedes principle, the loss of weight is equal to the weight of the displaced liquid. In the problem the liquid is the water whose density is 1 g/cm^3.
Hence when the sculpture is immersed in water, the new tension is
T' = mg - mg*ρ(w)/ρ(Al)
The frequency is directly proportional to the square root of the tension. Hence
f/f' = sqrt(T/T')
250/f' = sqrt[1/(1 - 1/2.7)]
Now solve for f'.
 
Last edited:
so in this equation, the 70kg value does not matter?

by the way, thanks a lot! :)
 
ngcg said:
so in this equation, the 70kg value does not matter?

by the way, thanks a lot! :)

Since we are taking the ratio of tensions, 70 kg does not matter.
 
thank you!
 

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