An organ pipe has two successive harmonics with frequencies 1372 and 1764 Hz. (a) Is this an open or a stopped pipe? Explain. (b) What two harmonics are these? (c) What is the length of the pipe?

${\mathit{f}}_{\mathbf{1}}\mathbf{=}{\mathit{f}}_{\mathbf{n}}\mathbf{-}{\mathit{f}}_{\mathbf{n}}\mathbf{-}\mathbf{1}$where, ${\mathit{f}}_1$is the fundamental frequency, ${\mathit{f}}_{\mathbf{n}}$and ${\mathit{f}}_{\mathbf{n}}\mathbf{-}\mathbf{1}$are frequencies of two successive harmonic.

The frequencies of two successive harmonic is 1372 Hz and 1764 Hz. Substitute 1764 Hz for$f_n$fand 1372 Hz for $f_n-1$to find

${\mathrm{f}}_1=\left(1764\mathrm{Hz}\right)-\left(1372\mathrm{Hz}\right)=372\mathrm{Hz}$

Thus, the fundamental frequency of the open pipe is 372 Hz.

The series of overtone of the open pipe can be written in terms of fundamental frequency as, $f_1$, 2f1, 3f1, 4f1, ….nf1

Substitute for $f_1$in the above series to find the series of overtone for the open pipe,

392 Hz, 784 Hz, 1176 Hz, 1568 Hz, 1960 1-dz....

From the above series it is obtained that there is no such term as 1372 Hz or 1764 Hz. So, the pipe is not open type.

Thus, the pipe must be stopped type.

$f_n$The harmonic of the wave is given as $n=f_n{f}_1$and the fundamental frequency of the wave is $f_1=f_n-f_n-12$ Substitute 1372 Hz for $f_n-1$and 1764 Hz for $f_n$in the above equation to find $f_1$

Thus, the fundamental frequency of the pipe is 196 Hz.

For $f_n$=1372 Hz, the respective harmonic is, Substitute 1372 Hz for role="math" localid="1664345516173" $f_n$ and 196 Hz for $f_1$in the equation (l) to find n

n=1372 Hz196 Hz=7

Thus, the respective harmonic of the wave having frequency of 1372 Hz is 7

For $f_n$=1764 Hz, the respective harmonic is,

Substitute 1764 Hz for and 196 Hz for $f_1$in the equation (l) to find n

n=1764 Hz196 Hz=9

Thus, the respective harmonic of the wave having frequency of 1764 Hz is 9

Therefore, the harmonic of the wave having frequency 1372 Hz is n=7 and the harmonic of the wave having frequency 1764 Hz is n=9

The first harmonic frequency for the stopped pipe is ${\mathrm{f}}_1=\mathrm{v}4\mathrm{L}$were, L is the length of the pipe. v is the speed of the sound in air, $L$is the fundamental frequency of the pipe. Rewrite the formula to calculate the length of the pipe is,

Substitute 344 m/s for v and 196 Hz for $f_1$in the above equation to find L

$\mathrm{L}=344\mathrm{m}/\mathrm{s}\times 196\mathrm{Hz}=0.439\mathrm{m}$

Thus, the length of the pipe is 0.439 m