Find the Second Resonant length of an air column

  • Thread starter Spookie71
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  • #1
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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
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
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.
 
  • #3
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Does that mean that both questions have the same answer?
 
  • #4
rl.bhat
Homework Helper
4,433
8
Yes.
 

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