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

Length of the pipe, $l=20cm=0.2m$

Source frequency $=n^{th}$ normal mode of frequency, $f_n=430Hz$
Speed of sound, v = 340 m/s
In a closed pipe, the $n^{th}$ normal mode of frequency is given by the relation:

$f=(2n-1)\frac{v}{4l}$ $n\in \{1,2,3,....\}$

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

$2n-1=\frac{430\times 4\times 0.2}{340}=1.01$

$2n=2.01$

$n\approx 1$

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:

$f_n=nv/2l$

$n=2l{f}_n/v\approx 0.5$

The same source will not be in resonance with the same pipe open at both ends.