Fundamental Frequency of an organ

  • #1

Homework Statement


An open organ pipe (i.e., a pipe open at both ends) of length L0 has a fundamental frequency f0.

Part A
If the organ pipe is cut in half, what is the new fundamental frequency?

4f0
2f0
f0
f0
f0

Part B
Part C

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Homework Equations



f=v/2L


The Attempt at a Solution



I am really confused by the standing waves and fundamental frequencies. The book does not do a good job explaining how this all works.

Anyways...for this individual problem I was thinking it might be 2f0.

if L is half as long, then the frequency is twice as big?
 

Answers and Replies

  • #2
Kurdt
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Sounds good to me.
 
  • #3
Thank you Kurdt. 2f0 was the correct answer.

Part B has revealed itself.

Part B
After being cut in half in Part A, the organ pipe is closed off at one end. What is the new fundamental frequency?


Homework Equations



f=v/2L


The Attempt at a Solution



The fundamental frequency of an open-closed tube is half that of an open-open or a closed-closed tube of the same length.

So...that means that the answer is f0/2?
 
  • #4
Kurdt
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Well be careful because remember the pipe was halved as well.
 
  • #5
Hm...so...

Cutting it in half made the frequency 2f0

Then making it open-closed...

2f0/2 = f0?
 
  • #6
Kurdt
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Yes. That seems fine.
 
  • #7
Part C
The air from the pipe in Part B (i.e., the original pipe after being cut in half and closed off at one end) is replaced with helium. (The speed of sound in helium is about three times faster than in air.). What is the approximate new fundamental frequency?

3f0
2f0
f0
f0/2
f0/3


I'm thinking the frequency gets bigger...so...3f0?

This is the last part of this question.
 
  • #8
Kurdt
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Yes that seems Ok too. :smile:
 

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