Acoustic Problem

  1. 1. The problem statement, all variables and given/known data
    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 ?


    2. Relevant equations



    3. 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: Nov 9, 2008
  2. jcsd
  3. Hootenanny

    Hootenanny 9,681
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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.
     
  4. Hootenanny

    Hootenanny 9,681
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    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)?
     
  5. Ohh Sorry !!

    For C Please take a look at my first post.
     
  6. Hootenanny

    Hootenanny 9,681
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    Yet you've done it again :grumpy:

    f = n*f1 looks like the right equation to apply.
     
  7. could you confirm me about the fundamental fr in (C)
     
  8. Hootenanny

    Hootenanny 9,681
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    f1 is given to you in the question:
     
  9. 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
     
  10. Hootenanny

    Hootenanny 9,681
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    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?
     


  11. The fundamental frequ. of an open pipe is

    f0 = v/(2*L1)
     
  12. Hootenanny

    Hootenanny 9,681
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    Correct, so using this equation and your previous one we can write

    [tex]f_n = n\frac{v}{2L}[/tex]

    Can you go from here?

    HINT: You know that the harmonics are consecutive.
     
  13. [tex]f_n = n\frac{v}{2L}[/tex]



    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 ?
     
    Last edited: Nov 9, 2008
  14. Hootenanny

    Hootenanny 9,681
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    Try writing them in the form

    [tex]f_n = n\frac{v}{2L}[/tex]

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

    [tex]f_{n+2} = \left(n+2\right)\frac{v}{2L}[/tex]
     
  15. ok lets 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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