
#1
Sep1908, 11:16 PM

P: 52

1. The problem statement, all variables and given/known data
An open organ pipe (i.e., a pipe open at both ends) of length L_{0} has a fundamental frequency f_{0}. Part A If the organ pipe is cut in half, what is the new fundamental frequency? 4f_{0} 2f_{0} f_{0} f_{0} f_{0} Part B Part C This part will be visible after you complete previous item(s). 2. Relevant equations f=v/2L 3. 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 2f_{0}. if L is half as long, then the frequency is twice as big? 



#2
Sep2008, 04:43 AM

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PF Gold
P: 4,975

Sounds good to me.




#3
Sep2008, 11:58 AM

P: 52

Thank you Kurdt. 2f_{0} 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? 2. Relevant equations f=v/2L 3. The attempt at a solution The fundamental frequency of an openclosed tube is half that of an openopen or a closedclosed tube of the same length. So...that means that the answer is f_{0}/2? 



#4
Sep2008, 12:09 PM

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PF Gold
P: 4,975

Fundamental Frequency
Well be careful because remember the pipe was halved as well.




#5
Sep2008, 12:16 PM

P: 52

Hm...so...
Cutting it in half made the frequency 2f_{0} Then making it openclosed... 2f_{0}/2 = f_{0}? 



#6
Sep2008, 12:22 PM

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PF Gold
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Yes. That seems fine.




#7
Sep2008, 12:39 PM

P: 52

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? 3f_{0} 2f_{0} f_{0} f_{0}/2 f_{0}/3 I'm thinking the frequency gets bigger...so...3f_{0}? This is the last part of this question. 



#8
Sep2008, 01:28 PM

Emeritus
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PF Gold
P: 4,975

Yes that seems Ok too.



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