# F frequency of small diameter organ pipe

1. Dec 17, 2005

### msimard8

This is a multiple choice question

The fundamental frequency of a small diameter organ pipe is

a) directly proportional to its length
b) inversely proportional to its length
c) independent of its length
d) inversely proportional to its diameter
e) directly proportional to its diameter

well this is what i know (assume)

the organ pipe is closed at one end

what formulas should i consider solving this problem

2. Dec 17, 2005

### mukundpa

go to text

3. Dec 17, 2005

### msimard8

whats that suppose to mean

4. Dec 17, 2005

### mukundpa

in your text book i think they have derived formulae for fundamental frequency and their overtones for open and cloced organ pipes, with the diagrams.

5. Dec 17, 2005

### msimard8

the only formula i see is

Ln=(2n=1) wavlength/4

which doesnt describe the question

Right now I am thinking that the diameter has no effect because, all diameter does is increase or decrease the amplitude of the wavelength which effects the loudness or intensity of the sound.

So therefore d and e are eliminated

The length of the pipe determines how many waves can fit in the pipe.

umm so confused

6. Dec 18, 2005

### mukundpa

dont get confused,
The formula is about the of a closed tube rasonating with nth harmonic of a given frequency f, and is
Ln=(2n-1) wavlength/4
( if it is + it is for nth overtone)
(n = 1) gives first hormonic or fundamentalfrequency,length is wavlength/4
(n = 2) gives first overtone frequency, length is 3*wavlength/4
(n = 3) gives second overtone frequency, length of tube is 5*wavlength/4 and so on
forget this here
Now on to the question
The frequency is given by
f = c/lembda = wave velocity / wavelength
in a colsed organ pipe standing waves will be produces with wavelength
4L(fundamental)= f0
4L/3(first overtone) = f1 = 3f0
5L/4(second overtone) =f2 =5f0 and so on
so the fundamental frequency of a close orgon pipe is
f0 = c/(4*L)

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