A pipe, 30.0 cm long. Is open at both ends. 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 m s^(-1)`.

Explanation

The first harmonic frequency is given by
`v_(1) = (v)/(lamda_(1)) = (v)/(2L)` (open pipe)
where L is the length of the pipe. The frequency of its nth harmonic is:
`v_(n) = (nv)/(2L)`, for n = 1,2,3. ……….(open pipe)
First few modes of an open pipe are shown in Fig. 15.15.


For L = 30.0 cm, `v = 330 m s^(-1)`,
`v_(n) = (n 330 (m s^(-1)))/(0.6 (m)) = 550 n s^(-1)`
Clearly , a source of frequency 1.1 kHz will resonate at `v_(2)`, i.e. the second harmonic.
Now if one of the pipe is closed (Fig. 151.15).
it follows from Eq. (14.50) that the fundamental frequency is
`v_(1) = (v)/(lamda_(1)) = (v)/(4L)` (pipe closed at one end)
and only the odd numnbered harmonics are present :
`v_(3) = (3v)/(4L), v_(5) = (5v)/(4L)`, and so on.
For L = 30 cm and `v = 330 ms^(-1)`, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.