A pipe 30.0 cm long is open at both ends. Which harmonic mode of the pipe resonates with a 1.1kHz 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 `330ms^(-1)`.

Explanation

Given, length of pipe, L = 30 cm = 0.30 m
Speed of sound, `upsilon` = 330 m/s
Fundamental frequency of the open pipe
`v_(1) = (upsilon)/(2L) = (330)/(2 xx 0.30) = 550` Hz
Second, third and fourth harmonics are
`{:(v_(2) = 2v_(1) = 2 xx 550 = 1100 Hz),(v_(3) = 3v_(1) = 3 xx 550 = 1650 Hz),(v_(4) = 4v_(1) = 4 xx 550 = 2200 Hz):}`
and so on.
Second harmonic of the open pipe will be resonantly excited by a source of frequency 1.1 kHz (or 1100 Hz).
If one end of the pipe is closed, its fundamental frequency is
`v._(1) = (upsilon)/(4L) = (330)/(4 xx 0.30) = 275` Hz
Since only odd harmonic are present in a closed pipe, so
Third harmonic, `v._(3) = 3v._(1) = 3 xx 275 = 825` Hz
Fifth harmonic, `v._(5) = 5v._(1) = 5 xx 275 = 1375` Hz
and so on.
We can see that no frequency of the closed pipe matches with the source frequency of 1.1 KHz, so no resonance can be observed with the source at the instant when the pipe is closed.