Question

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

Answer 1

Given,

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

Source frequency$=n^{th}$ normal mode of frequency,$v_n=430Hz$

Speed of sound, $v=340m/s$

Step 1: Find the mode of frequency when pipe is closed

In a closed pipe, the$n^{th}$ mode of frequency is given by,

$\nu =\frac{(n+\frac{1}{2})v}{2l}$

Now for the given conditions,

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

$\Rightarrow n+\frac{1}{2}=\frac{430\times 0.4}{340}$

$\Rightarrow n+\frac{1}{2}\approx \frac{1}{2}\Rightarrow n=0$

Here, the fundamental mode of vibration frequency is resonantly excited by the given source.

Step 2: Find the mode of frequency when pipe is open

Now when the pipe is open at both ends, the $n^{th}$ mode of vibration frequency is given by,

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

For the given conditions,

$430=\frac{n\times 340}{2\times 0.2}$

$\Rightarrow n=\frac{430\times 0.4}{340}\approx \frac{1}{2}$

Since the n has to be an integer, the given source does not produce a resonant vibration in an open pipe.