Relating resonance with two different Pipes

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SUMMARY

The discussion centers on determining the length of pipe A, which is open at one end, such that its third lowest resonant frequency matches the lowest resonant frequency of pipe B, which is open at both ends. The relevant equations include the frequency formulas for the harmonics of each pipe: for pipe A, the fifth harmonic is given by f = (5v)/(4L), and for pipe B, the first harmonic is f = v/2L. The solution requires setting these two frequency equations equal to each other to find the appropriate length of pipe A.

PREREQUISITES
  • Understanding of wave mechanics and resonance
  • Familiarity with harmonic frequencies in open and closed pipes
  • Knowledge of the wave equation v = (lambda)(f)
  • Basic algebra for solving equations
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  • Study the principles of resonance in open and closed pipes
  • Learn about harmonic frequencies and their calculations
  • Explore the wave equation and its applications in acoustics
  • Practice solving problems involving resonant frequencies of different pipe configurations
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Students in physics, particularly those studying acoustics and wave mechanics, as well as educators seeking to explain the concepts of resonance in pipes.

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Homework Statement



A pipe, open at one end, has a leng
produces pipe A, open at one end, and pipe B, o

A pipe, open at one end, has a length L. Cutting this pipe crosswise
produces pipe A, open at one end, and pipe B, open at both ends.
a. For what length of pipe A will the third lowest resonant frequency of pipe A be equal
to the lowest resonant frequency of pipe B?


Homework Equations



v = (lambda)(f)

5th harmonic of A, f = (5v)/(4L)
1st Harmonic of B, f = v/2L

The Attempt at a Solution



I've been fairly stumped by this one. I don't get exactly how to set it up. I tried setting the frequencies equal to each other, but it doesn't work.
 
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I'm really just looking for a hint to set me on the right path.
 

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