# Fundamental Frequency

by Foxhound101
Tags: frequency, fundamental
 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 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 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 2f0. if L is half as long, then the frequency is twice as big?
 Emeritus Sci Advisor PF Gold P: 4,975 Sounds good to me.
 P: 52 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? 2. Relevant equations f=v/2L 3. 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?
Emeritus
PF Gold
P: 4,975

## Fundamental Frequency

Well be careful because remember the pipe was halved as well.
 P: 52 Hm...so... Cutting it in half made the frequency 2f0 Then making it open-closed... 2f0/2 = f0?
 Emeritus Sci Advisor PF Gold P: 4,975 Yes. That seems fine.
 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? 3f0 2f0 f0 f0/2 f0/3 I'm thinking the frequency gets bigger...so...3f0? This is the last part of this question.
 Emeritus Sci Advisor PF Gold P: 4,975 Yes that seems Ok too.

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