A TALE OF TWO PIPES
On a day when the speed of sound is 345 m/s, the fundamental frequency of a particular stopped organ pipe is 220 Hz. (a) How long is this pipe? (b) The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe. How long is the open pipe?
IDENTIFY and SET UP:
This problem uses the relationship between the length and normal-mode frequencies of open pipes (Fig. 16.17) and stopped pipes (Fig. 16.18). In part (a), we determine the length of the stopped pipe from Eq. (16.22). In part (b), we must determine the
length of an open pipe, for which Eq. (16.18) gives the frequencies.
f_{n}=\frac{n v}{2 L}Â Â Â Â Â Â Â Â Â Â Â Â (16.18)
f_{n}=\frac{n v}{4 L}Â Â Â Â Â Â Â Â Â Â Â Â (16.22)
EXECUTE:
(a) For a stopped pipe f_1 = v/4L, so
L_{\text {stopped }}=\frac{v}{4 f_{1}}=\frac{345 \mathrm{~m} / \mathrm{s}}{4\left(220 \mathrm{~s}^{-1}\right)}=0.392 \mathrm{~m}(b) The frequency of the second overtone of a stopped pipe (the third possible frequency) is f_5 = 5f_1 = 5(220 Hz) = 1100 Hz. If the wavelengths for the two pipes are the same, the frequencies are also the same. Hence the frequency of the third harmonic of the open pipe, which is at 3f_1 = 3(v/2L), equals 1100 Hz. Then
1100 \mathrm{~Hz}=3\left(\frac{345 \mathrm{~m} / \mathrm{s}}{2 L_{\mathrm{open}}}\right) \quad \text { and } \quad L_{\mathrm{open}}=0.470 \mathrm{~m}
EVALUATE: The 0.392-m stopped pipe has a fundamental frequency of 220 Hz; the longer (0.470-m) open pipe has a higher fundamental frequency, (1100 Hz)/3 = 367 Hz. This is not a contradiction, as you can see if you compare Figs. 16.17a and 16.18a.
