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).

A pipe, 30.0 cm long. Is open at both ends. Which harmonic mode of the

1.	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).
Answer» Solution :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.