Resonant Lengths in open air column question

  • #1
3
0

Homework Statement


An organ pipe 1.2m long and open at both ends produces a note with the fundamental frequency. If the speed of sound in air is 345 m/s, what is the fundamental frequency?

Homework Equations


Wave equation (f = v/lambda)

The Attempt at a Solution


My textbook solves the problem like so:
6463fc868ce2f824d3fc1f9a40795ea2.png


My question is: why do they use the full wavelength here? As I understand it, the fundamental frequency is only half a wavelength. This would be more like the second resonant length, which was not what the question asks.
 

Attachments

  • 6463fc868ce2f824d3fc1f9a40795ea2.png
    6463fc868ce2f824d3fc1f9a40795ea2.png
    13 KB · Views: 697

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
37,988
7,702
the fundamental frequency is only half a wavelength
First, a frequency is not a length.
Fundamental frequency means the lowest note the pipe can produce. Being open at both ends, the pipe will only contain half a wavelength. Thus, as you imply, the fundamental frequency here corresponds to a half wavelength. Hence the λ=2L.
But to find the frequency that you hear you must divide the speed of sound by the full wavelength.
 
  • #3
3
0
First, a frequency is not a length.
Fundamental frequency means the lowest note the pipe can produce. Being open at both ends, the pipe will only contain half a wavelength. Thus, as you imply, the fundamental frequency here corresponds to a half wavelength. Hence the λ=2L.
But to find the frequency that you hear you must divide the speed of sound by the full wavelength.
Thanks for your reply.

So the wavelength has to be twice the length of the pipe because the pipe is open on both sides? So the fundamental frequency corresponds to the actual length of the pipe?

How would this be solved if the pipe were closed on one end?
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
37,988
7,702
How would this be solved if the pipe were closed on one end?
In that case, how much of a wavelength (at fundamental frequency) would fit in the pipe?
 
  • #5
3
0
In that case, how much of a wavelength (at fundamental frequency) would fit in the pipe?

1/4 wavelength?

Also, was the rest of the conclusion earlier correct?
 
  • #7
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
37,988
7,702
the fundamental frequency corresponds to the actual length of the pipe?
It might be clearer to think in terms of the "fundamental wavelength", i.e. the longest wavelength which can be produced. That is related, on the one hand, to the pipe length and its number of closed ends, and on the other to the fundamental frequency by the speed of sound in air.
 

Related Threads on Resonant Lengths in open air column question

Replies
4
Views
6K
  • Last Post
Replies
3
Views
9K
Replies
4
Views
3K
Replies
3
Views
21K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
4K
Replies
0
Views
6K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
6
Views
763
Top