Question

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source ? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s-1).

Answer 1

First (Fundamental); No
Length of the pipe, $l$ = $20 \;cm$ = $0.2 \;m$
Source frequency = $n^{th}$ normal mode of frequency is $V_n$ = $430 \;Hz$
Speed of sound, $v$ = $340 \;m/s$
In a closed pipe, the $n^{th}$ normal mode of frequency is given by the relation:
$v_n$ = $(2n-1)\frac{v}{4l}$ ; $n$ is an interger = $0,1,2,3,4$

$430$ = $(2n-1)\frac{340}{4\times 0.2}$

$2n-1$= $\frac{430\times 4\times 0.2}{340}$ = $1.01$
$2n$ = $2.01$
$n∼1$
Hence, the first mode of vibration frequency is resonantly excited by the given source.
In a pipe open at both ends, the nth mode of vibration frequency is given by the relation

$v_n$ = $\frac{nv}{2l}$

$n$ = $\frac{2lV_n}{v}$

= $\frac{2×0.2×430}{340}$ = $0.5$

Since the number of the mode of vibration ($n$) has to be an integer, the given source does not produce a resonant vibration in an open pipe.