P18_040.PDF

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Chapter 18 - 18.40
40. We observe that “third lowest ... frequency” corresponds to harmonic number n = 3 for a pipe open at
both ends. Also, “second lowest ... frequency” corresponds to harmonic number n = 3 for a pipe closed
at one end.
(a) Since λ =2 L/n for pipe A ,where L =1 . 2m,then λ =0 . 80 m for this mode. The change from
node to antinode requires a distance of λ/ 4 so that every increment of 0 . 20 m along the x axis
involves a switch between node and antinode. Since the opening is a displacement antinode, then
the locations for displacement nodes are at x =0 . 20 m, x =0 . 60 m, and x =1 . 0m.
(b) The waves in both pipes have the same wavespeed (sound in air) and frequency, so the standing
waves in both pipes have the same wavelength (0 . 80 m). Therefore, using Eq. 18-38 for pipe B ,we
find L =3 λ/ 4=0 . 60 m.
(c) Using v = 343 m/s, we find f 3 = v/λ = 429 Hz. Now, we find the fundamental resonant frequency
by dividing by the harmonic number, f 1 = f 3 / 3 = 143 Hz.

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