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Worked Examples · Example 14.5

Q.A pipe, 30.0 cm30.0\ \text{cm} long, is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz1.1\ \text{kHz} source? Will resonance with the same source be observed if one end of the pipe is closed? Take the speed of sound in air as 330 m s−1330\ \text{m s}^{-1}.

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For the 30.0 cm30.0\ \text{cm} pipe open at both ends, the fundamental is 550 Hz550\ \text{Hz}, and the 1.1 kHz1.1\ \text{kHz} source matches its second harmonic (n=2n=2). If one end is closed instead, only odd harmonics of the (now lower) fundamental 275 Hz275\ \text{Hz} are allowed, and 1100 Hz1100\ \text{Hz} is an even multiple of that -- so no resonance occurs with one end closed.

Figure 14.15
Figure 14.15

Open pipe (both ends open)

For a pipe open at both ends, both ends are displacement antinodes, and the allowed resonant frequencies are all integer multiples of the fundamental:

fn=nv2L,n=1,2,3,…f_n = \frac{nv}{2L}, \qquad n=1,2,3,\dots

Step 1 -- fundamental frequency.

f1=v2L=3302×0.300=550 Hzf_1 = \frac{v}{2L} = \frac{330}{2\times0.300} = 550\ \text{Hz}

Step 2 -- which harmonic matches 1.1 kHz?

n=fsf1=1100550=2n = \frac{f_s}{f_1} = \frac{1100}{550} = 2

So the source resonates with the pipe's second harmonic.

Same pipe, one end closed

For a pipe closed at one end, the closed end is a displacement node and the open end is a displacement antinode. This asymmetric boundary condition only permits odd harmonics of a different, lower fundamental:

fn′=nv4L,n=1,3,5,…f_n' = \frac{nv}{4L}, \qquad n=1,3,5,\dots

Step 3 -- new fundamental.

f1′=v4L=3304×0.300=275 Hzf_1' = \frac{v}{4L} = \frac{330}{4\times0.300} = 275\ \text{Hz}

Allowed resonant frequencies: 275, 825, 1375,… Hz275,\ 825,\ 1375,\dots\ \text{Hz} (odd multiples of 275275 only).

Step 4 -- check whether 1100 Hz appears.

1100275=4\frac{1100}{275} = 4 …

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