Question

Two adjacent natural frequencies of an organ pipe are determined to be 780 Hz and 884 Hz. (Assume the speed of sound is 343 m/s.)

(a) Calculate the fundamental frequency of this pipe.
Hz

(b) Calculate the length of this pipe.
m

Answer 1

(A) For an open pipe the fundamental frequency Fo = V/2L where L is the length of pipe and v is velocity of sound in air. The overtones are thus 2Fo, 3Fo,4Fo etc.

For a closed pipe the fundamental frequency Fo =V/4L and the overtones are 3Fo, 5Fo,7Fo, etc.

If the pipe is open the nth harmonic would be

nFo = 780

and the next harmonic would be

(n+1)Fo =884

subtract these and Fo = 104, but with Fo it is not possible to have frequencies given so it must be a closed pipe.

i.e nFo = 780 and (n+2) =884

i.e 2Fo = 104 or Fo = 52 Hz

(B) so 52 = 343/4L i.e L = 343/(52*4) = 1.65 m