A pipe , 30.0 cm long is open at bothends. Which harmonic mode of the pipe resonates a 1.1 kHz source ?Will resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as `330 ms^(-1)`

Explanation

The first harmonic frequency is given by `v^(1) = (v )/( lambda_(1))= ( v)/( 2L) ` ( open pipe)
When L is the length of the pipe.The frequency of its `n^(th)` harmonic is
`vn = (nv)/( 2L) ` for n = 1,2,2, …...(open pipe)
First few modes of an open pipe are shown in Fig.
For L ` =30.0cm., v= 330 ms^(-1)`
`v_(n )=(n xx330 ( ms^(-1)))/(0.6(m))=550s^(-1)`

Clearly, for a source of frequency 1.1kHz the air column will resonate at `v_(2)` i.e.,the second haremonic .
Nowif one end of the pipe is closed (Figure) the fundamental frequency is
`v_(1) = (v)/( lambda_(1))= (v)/( 4L)` ( pipe closed at one end )

and only to odd numbered harmonics are present `:`
`v_(3) =( 3v)/( 4L) , v_(5) =( 5v)/( 4L)`
For L = 30cm and v `= 330 m s^(-1)` , the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its forth harmonic. Since this harmonic is not a possible mode,no resonance will be observed with the source, the moment one end is closed.