A pipe closed at one end vibrates in its first overtone. An open pipe vibrating in its fun… — JAMB UTME

A pipe closed at one end vibrates in its first overtone. An open pipe vibrating in its fundamental mode has a frequency equal to that of the closed pipe. Neglecting end corrections, what is the ratio of the length of the open pipe to the length of the closed pipe?

$2 : 3$ Answer
B
$3 : 2$
C
$4 : 3$
D
$1 : 3$

Answer

The ratio of the length of the open pipe to the length of the closed pipe is

2 : 3

The first overtone of a closed pipe corresponds to its 3rd harmonic, giving a frequency of

f = \frac{3v}{4L_c}

. Equating this to the fundamental frequency of an open pipe (

f = \frac{v}{2L_o}

) yields

\frac{v}{2L_o} = \frac{3v}{4L_c}

, which simplifies to

\frac{L_o}{L_c} = \frac{2}{3}

2 : 3

Step-by-Step Solution

Write the frequency formula for the first overtone of the closed pipe.

For a closed pipe of length

L_c

, odd harmonics are produced (

n = 1, 3, 5, \dots

). The first overtone is the third harmonic (

n = 3

f_{\text{closed}} = \frac{3v}{4L_c}

Closed air columns produce only odd harmonics, where the fundamental is

n=1

and the first overtone is

n=3

Write the fundamental frequency formula for the open pipe.

For an open pipe of length

L_o

, all harmonics are produced (

m = 1, 2, 3, \dots

). The fundamental frequency (

m = 1

) is:

f_{\text{open}} = \frac{v}{2L_o}

Open air columns have antinodes at both ends, yielding a fundamental wavelength of

\lambda = 2L_o

Equate the two frequencies and solve for the ratio

\frac{L_o}{L_c}

\frac{v}{2L_o} = \frac{3v}{4L_c}

Cancel the speed of sound

v

from both sides:

\frac{1}{2L_o} = \frac{3}{4L_c}

Cross-multiply:

6L_o = 4L_c \implies \frac{L_o}{L_c} = \frac{4}{6} = \frac{2}{3}

The question states that the frequencies of the two pipe configurations are equal.

Key Concept

Harmonics in Open and Closed Air Columns

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