Solving Closed Pipe Question: Length Difference of 2 Pipes

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SUMMARY

The discussion centers on a physics problem involving closed organ pipes, specifically a pipe of length 3.2 m producing a frequency of 27 Hz. The presence of a 1.50 Hz beat frequency indicates a difference in frequency between the two pipes. The solution requires calculating the frequency of the second pipe and determining its length based on the relationship between frequency and pipe length. The correct approach involves using the formula for the fundamental frequency of a closed pipe, which is given by f = v / (4L), where v is the speed of sound.

PREREQUISITES
  • Understanding of wave mechanics and sound frequency
  • Knowledge of the relationship between frequency and pipe length in closed organ pipes
  • Familiarity with the concept of beat frequency in acoustics
  • Basic algebra for solving equations
NEXT STEPS
  • Study the fundamental frequency formula for closed pipes: f = v / (4L)
  • Learn about beat frequency and its calculation in acoustics
  • Explore the speed of sound in air and its dependency on temperature
  • Practice problems involving multiple sound sources and interference patterns
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics and sound, as well as educators seeking to explain concepts related to organ pipes and acoustics.

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Homework Statement


Once closed organ pipe has a length of 3.2 m. Frequency of the note played by this pipe is 27 Hz. When a second pipe is played at the same time, a 1.50 Hz beat note is heard. By how much is the second pipe too long?


Homework Equations



?

The Attempt at a Solution



I'm really lost on this problem. I was thinking Lb-La=1/2 of wavelength, but I don't think that's right.

Any direction is appreciated. Thanks in advance.
 
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