Length of Wavelength in closed aircolumn

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SUMMARY

The discussion focuses on calculating the wavelength of the first overtone in an 80-cm long closed air column. The fundamental frequency wavelength is established as four times the length of the pipe, leading to the formula λ = 4L. For the first overtone in a closed air column, the wavelength is determined to be 4/3 of the length of the pipe, resulting in a wavelength of approximately 106.67 cm. The closed end of the pipe serves as a node, while the open end acts as an anti-node, confirming the standing wave pattern.

PREREQUISITES
  • Understanding of wave mechanics, specifically standing waves
  • Familiarity with the concepts of nodes and anti-nodes
  • Knowledge of the relationship between frequency, wavelength, and wave speed (v = fλ)
  • Basic principles of acoustics related to closed air columns
NEXT STEPS
  • Study the harmonic series in closed air columns
  • Learn about the impact of pipe diameter on wavelength calculations
  • Explore the differences between open and closed pipe harmonics
  • Investigate practical applications of standing waves in musical instruments
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics, acoustics, and sound engineering. This discussion is also beneficial for educators teaching concepts related to standing waves in closed air columns.

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Homework Statement



An 80-cm long air column, closed at one end, contains sounding waves. How long is the wavelength that corresponds to: the first overtune

Homework Equations



v = f(lambda)

The Attempt at a Solution



no clue about what to do.
 
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In a pipe with one closed end, the first harmonic (fundamental frequency) has a wavelength that is equal to 4 times the length of the pipe. This is because the closed end acts as as a node and the open end as an anti-node. What will the wavelength of the first overtone then be?
 
For a closed air column, the most simple standing wave that can occur is one with a nodal position at the closed end and an antinodal position at the open end...or 1/4 of a complete wavelength would correspond to the length of the column. Thus lambda = 4L.
There is a correction factor for the diameter of the pipe, but since you did not include that information, I will assume that the small correction factor is being ignored here.


Edit to Add...Kurdt beat me to it!
 

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