A pipe closed at one end vibrates in its fundamental mode with a frequency equal to that o… — JAMB UTME

A pipe closed at one end vibrates in its fundamental mode with a frequency equal to that of an open pipe vibrating in its first overtone. What is the ratio of the length of the closed pipe to the length of the open pipe?

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

Answer

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

1 : 4

The fundamental mode of a pipe closed at one end has a frequency of

f = \frac{v}{4L_c}

. The first overtone of a pipe open at both ends corresponds to the second harmonic, which has a frequency of

f = \frac{v}{L_o}

. Equating the two frequencies gives

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

, which simplifies directly to

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

1 : 4

Step-by-Step Solution

Express the fundamental frequency of the pipe closed at one end

f_c = \frac{v}{4L_c}

, where

v

is the speed of sound and

L_c

is the length of the closed pipe.

A pipe closed at one end supports odd harmonics, and its fundamental wavelength is

\lambda_c = 4L_c

Express the frequency of the first overtone of the open pipe

f_o = \frac{2v}{2L_o} = \frac{v}{L_o}

, where

L_o

is the length of the open pipe.

An open pipe supports all harmonics (

n = 1, 2, 3, \dots

). The fundamental is

n=1

and the first overtone corresponds to

n=2

Equate the two frequencies and solve for the length ratio

\frac{L_c}{L_o}

\frac{v}{4L_c} = \frac{v}{L_o} \implies 4L_c = L_o \implies \frac{L_c}{L_o} = \frac{1}{4}

The problem states that the fundamental frequency of the closed pipe is equal to the first overtone frequency of the open pipe.

Key Concept

Standing Waves and Harmonics in Pipes

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