Find the Second Resonant length of an air column

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Homework Help Overview

The problem involves determining the second resonant length of an air column that resonates with a frequency of 1.0 kHz at 15.0 degrees Celsius, under two different conditions: one where the air column is closed at both ends and another where it is open at both ends.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the second resonant length using the speed of sound and wavelength formulas, but expresses uncertainty about the differences in calculations for closed and open air columns. Other participants discuss the nature of resonant lengths in relation to nodes and antinodes in both types of columns.

Discussion Status

Participants are exploring the implications of the calculations for both types of air columns. Some guidance has been offered regarding the relationship between resonant lengths and the characteristics of nodes and antinodes, but there is still some uncertainty about whether the answers for both conditions are indeed the same.

Contextual Notes

The original poster notes a lack of clarity in their textbook regarding the differences in calculations for the two types of air columns, which may affect their understanding of the problem.

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Homework Statement


Find the second resonant length of an air column that resonates with a sound of frequency 1.0 kHz at 15.0 degrees Celsius under each of the following conditions.

a) the air column is closed at both ends

b) the air column is open at both ends

Homework Equations


[tex]V_{s}[/tex] = 332 m/s + T(0.59 m/s [tex]\circ[/tex]C)

v = f[tex]\lambda[/tex] therefore [tex]\lambda[/tex] = [tex]\frac{v}{f}[/tex]

l = [tex]\frac{n \lambda}{2}[/tex] therefore [tex]\frac{2l}{n}[/tex] = [tex]\lambda[/tex]

n = 2 for the second resonant length of an air column


The Attempt at a Solution


[tex]V_{s}[/tex] = 332 m/s + T(0.59 m/s [tex]\circ[/tex]C)
332 m/s + 15(0.59 m/s)
= 340.85
-------------------------------------------------------------
v = f[tex]\lambda[/tex] therefore [tex]\lambda[/tex] = [tex]\frac{v}{f}[/tex]

[tex]\frac{340.85 m/s}{1000 Hz}[/tex]
= 34.09 cm
-------------------------------------------------------------

[tex]\frac{2l}{n}[/tex] = [tex]\lambda[/tex]

[tex]\lambda[/tex] = [tex]\frac{2(34.09}{2}[/tex]
= 34.09

--------------------------------------------------------

The second Resonant length is 34.09 cm
I don't know how to calculate the difference between an air column open at both ends
and an air column closed at both ends. My textbook doesn't explain it clearly. I'm guessing that both types of columns have different answers but as it stands I got the same calculation for both types. Is there more to the equation that I'm missing or am I doing the whole calculation wrong.

Thanks
S
 
Physics news on Phys.org
In open air column open ends will be antinodes. So the second resonant length = wavelength. In closed air column closed ends will be nodes. So the second resonant length is also equal to wavelength.
 
Does that mean that both questions have the same answer?
 
Yes.
 

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