D)We can use the same formula as in part A. L = 343 m.s / 466.2 HzL = 0.737m

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The discussion focuses on calculating the lengths and harmonics of organ pipes, specifically one open at both ends and one closed at one end, both tuned to a fundamental frequency of 264 Hz. The calculations reveal that the length of the open pipe is 0.649 meters and the closed pipe is 0.3247 meters. The frequencies of three neighboring harmonics (466.2 Hz, 582.7 Hz, and 699.2 Hz) correspond to the 4th, 5th, and 6th harmonics of the open pipe, with the length of the pipe determined to be 1.471 meters. The fundamental frequency equation for open pipes is confirmed as f0 = v/(2L).

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Homework Statement


One organ pipe open on both ends and one organ pipe open at one end and closed
at the other are tuned on the same fundamental tone with frequency f = 264 Hz.

a)How long is each pipe?
b) Determine the first three harmonics of each pipe.
c) There is an organ pipe open at both ends, where the frequencies of three neighbo-
ring harmonics are 466:2 Hz, 582:7 Hz and 699:2 Hz. To which harmonics do this
frequencies belong?
d) How long is this pipe?
Velocity c = 343 m/s.whats is the basic difference if an organ pipe in open on both ends or one end closed ?
To find the length in part d) should i apply the same relationship ?

Homework Equations


The Attempt at a Solution



I tried like ,A)
Since Wavelength * Freq = Speed
Therefore Wavelength = Speed / Freq
Wavelength = 343 m.s / 264 Hz
Wavelength =1.299m

For Open Pipe:
L = 1/2 (wavelength)
L = 0.649 meter

For Closed Pipe:
L= 1/4 Wavelength
L = 0.3247 meter
B) Open Pipe Closed Pipe
L = 1/2 Wavelength L = 1/4 Wavelength
L= 2/2 Wavelength L = 3/4 Wavelength
L= 3/2 Wavelength L = 5/4 Wavelength C) Should I apply fn = n f1 if yes so is the f1 = 264 Hz ?
 
Last edited:
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Closed and open pipes have different fundamental frequencies. The equation that you used are for open pipes only (L=λ/2), since a closed pipe has a different fundamental frequency you are required to use a different equation to determine the length of the pipes.
 
Please don't edit your post once someone has replied. Doing so can make their reply seem wrong or confusing, as in this case.

Which equations are you using for (c)?
 
Ohh Sorry !

For C Please take a look at my first post.
 
Hootenanny said:
Please don't edit your post once someone has replied. Doing so can make their reply seem wrong or confusing, as in this case.
SEEDS said:
Ohh Sorry !

For C Please take a look at my first post.
Yet you've done it again

f = n*f1 looks like the right equation to apply.
 
could you confirm me about the fundamental fr in (C)
 
SEEDS said:
could you confirm me about the fundamental fr in (C)
f1 is given to you in the question:
SEEDS said:
One organ pipe open on both ends and one organ pipe open at one end and closed
at the other are tuned on the same fundamental tone with frequency f = 264 Hz.
 
If I am using
fn = n f1 so i am getting fractional figures

466.2 / 264 = n and i guess n should'nt be fractional .
582.7 / 264 = n
699.2 / 264 = n
 
SEEDS said:
If I am using
fn = n f1 so i am getting fractional figures

466.2 / 264 = n and i guess n should'nt be fractional .
582.7 / 264 = n
699.2 / 264 = n
Hmm, indeed you are correct. Therefore, we must conclude that we are dealing with a different pipe to the one in the previous question. So, let us start again.

What is the general equation for the fundamental frequency of an open pipe?
 
  • #10
Hootenanny said:
What is the general equation for the fundamental frequency of an open pipe?



The fundamental frequ. of an open pipe is

f0 = v/(2*L1)
 
  • #11
SEEDS said:
The fundamental frequ. of an open pipe is

f0 = v/(2*L1)
Correct, so using this equation and your previous one we can write

f_n = n\frac{v}{2L}

Can you go from here?

HINT: You know that the harmonics are consecutive.
 
  • #12
f_n = n\frac{v}{2L}
I am applying it like

n = (fn * 2 L) / v but
n1=2.718 L
n2=3.397 L
n3=4.076 Lstill confusion how to find L to proceed finally for the harmonics ?
 
Last edited:
  • #13
SEEDS said:
f_n = n\frac{v}{2L}



I am applying it like

n = (fn * 2 L) / v but
n1=2.718 L
n2=3.397 L
n3=4.076 L


still confusion how to find L to proceed finally for the harmonics ?
Try writing them in the form

f_n = n\frac{v}{2L}

f_{n+1} = \left(n+1\right)\frac{v}{2L}

f_{n+2} = \left(n+2\right)\frac{v}{2L}
 
  • #14
ok let's check it

fn = n v/2L ---A

fn+1 = (n+1) v/2L ----B

fn = 466.2 Hz , fn+1 = 582.7 Hz

Therefore A becomes

466.2 = n (343 / 2L )
2.718L = n ------(i)

and B becomes by fn+1
582.7L = 171.5 n + 171.5 ---(ii)
putting n from i

L = 1.471 meter
and n is approx = 4
So the harmonics are 4th , 5th , 6th

Is this correct ?
 

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