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 ???
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)?
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?
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.
[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 ?
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]
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 ?