Fundamental frequency and changes to it

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SUMMARY

The fundamental frequency of a stretched wire is determined by the formula F = (1/2L)√(T/μ). In this discussion, a wire vibrating at 384 Hz is analyzed under three changes: halving its length, doubling its diameter, and increasing its tension five-fold. The new frequency can be calculated using the modified equation F2 = (1/2(L/2))√(5T/(4μ)), leading to the conclusion that the new fundamental frequency is 384√5 Hz.

PREREQUISITES
  • Understanding of wave mechanics and frequency calculations
  • Familiarity with the concepts of tension and mass per unit length (μ)
  • Knowledge of the relationship between physical dimensions and frequency in vibrating strings
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the effects of tension on wave speed in strings
  • Learn about the relationship between diameter and mass per unit length (μ) in cylindrical objects
  • Explore the concept of harmonics in vibrating strings
  • Investigate practical applications of frequency changes in musical instruments
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Physics students, educators, and anyone interested in the principles of wave mechanics and their applications in real-world scenarios such as musical acoustics.

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Homework Statement



A stretched wire vibrates in its fundamental mode at a frequency of 384 Hz. What would be the fundamental frequency if the wire were half as long, its diameter were doubled, and its tension were increased five-fold?

Homework Equations



F= [tex]\frac{1}{2L}[/tex][tex]\sqrt[]{\frac{T}{\mu}}[/tex]

The Attempt at a Solution


ok i know that this F= [tex]\frac{1}{2L}[/tex][tex]\sqrt[]{\frac{T}{\mu}}[/tex] is the formula for the one where F1= 384 Hz

i know that length is cut in half and that tension increases by 5and when the diameter is doubled that means mass is quadrupled my problem is that i don't know how to show that wrt [tex]\mu[/tex]
for the second frequency would my equation look like
F2= [tex]\frac{1}{2(L/2)}[/tex][tex]\sqrt[]{\frac{5*T}{4*\mu}}[/tex]
and then if we cancel off the 2 in the l would it look like F= [tex]\frac{1}{L}[/tex][tex]\sqrt[]{\frac{5*T}{\mu}}[/tex] *[tex]\frac{1}{2}[/tex]
 
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Looks like you have it then right?

384*√5
 
thanx i wasnt sure if it looked right
 

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