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.