In Acoustics a closed pipe or organ pipe means a tube closed at one end. An open pipe or organ pipe means a tube open at both ends. When a standing wave (stationary wave) is formed in an organ pipe, the closed end will be a node and the open end will be an antinode. This is why the length of the pipe in the fundamental mode is equal to λ/4 (which is the distance between neighbouring node and antinode) in a closed pipe. In an open pipe, in the fundamental mode, the length of the pipe is equal to λ/2 since the consecutive antinodes are located at the ends of the tube.
You should remember that a closed pipe can support odd harmonics only where as an open pipe can support all harmonics. In other words, the frequencies of vibration of the air column in a closed pipe are in the ratio 1: 3 : 5 : 7 : etc., while those in an open pipe are in the ratio 1 : 2 : 3 : 4 : 5 : etc.
Now, consider the following MCQ:
An open organ pipe and a closed organ pipe have the same length. The ratio of their fundamental frequencies is
(a) 1 (b) 2 (c) 3 (d) 4 (e) 3/4
If ‘L’ is the length of the pipe, we have, L = λ/2 for the open pipe and L = λ'/4 for the closed pipe where λ and λ' are the wave lengths of sound in the two cases (in the fundamental mode).
The corresponding fundamental frequencies are n = v/ λ = v/2L and n' = v/λ' = v/4L, from which n/n' = 2 [Option (b)].
Let us consider another question:
Almost the entire length of an aluminium pipe of length 1.1m is dipped vertically in water contained in a tall jar. An excited tuning fork of frequency 500 Hz is held at the upper end of the pipe and the pipe is gradually raised. How many discrete resonance conditions are possible? (Speed of sound in air = 330 m/s)
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
The arrangement mentioned in this problem makes a simple resonance column apparatus. The wave length of sound emitted by the fork, λ = v/n = 330/500 = 0.6666 m.
The first resonance (fundamental mode) is obtained when the exposed length of the pipe is λ/4. The second resonance is obtained when the exposed length is 3 λ/4. These two are definitely possible since the length of the pipe is 1.1 m and λ = 0.6666 m.
The third resonance will be obtained when the exposed length is 5λ/4 = 5×0.6666/4 = 0.83 m. This too is possible.
The fourth resonance will be obtained when the exposed length of the pipe is 7λ/4 = 7×0.6666/4 = 1.16 m. This is not possible since the length of the entire pipe is 1.1 m only.
So, the correct option is (c).
You should remember that a closed pipe can support odd harmonics only where as an open pipe can support all harmonics. In other words, the frequencies of vibration of the air column in a closed pipe are in the ratio 1: 3 : 5 : 7 : etc., while those in an open pipe are in the ratio 1 : 2 : 3 : 4 : 5 : etc.
Now, consider the following MCQ:
An open organ pipe and a closed organ pipe have the same length. The ratio of their fundamental frequencies is
(a) 1 (b) 2 (c) 3 (d) 4 (e) 3/4
If ‘L’ is the length of the pipe, we have, L = λ/2 for the open pipe and L = λ'/4 for the closed pipe where λ and λ' are the wave lengths of sound in the two cases (in the fundamental mode).
The corresponding fundamental frequencies are n = v/ λ = v/2L and n' = v/λ' = v/4L, from which n/n' = 2 [Option (b)].
Let us consider another question:
Almost the entire length of an aluminium pipe of length 1.1m is dipped vertically in water contained in a tall jar. An excited tuning fork of frequency 500 Hz is held at the upper end of the pipe and the pipe is gradually raised. How many discrete resonance conditions are possible? (Speed of sound in air = 330 m/s)
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
The arrangement mentioned in this problem makes a simple resonance column apparatus. The wave length of sound emitted by the fork, λ = v/n = 330/500 = 0.6666 m.
The first resonance (fundamental mode) is obtained when the exposed length of the pipe is λ/4. The second resonance is obtained when the exposed length is 3 λ/4. These two are definitely possible since the length of the pipe is 1.1 m and λ = 0.6666 m.
The third resonance will be obtained when the exposed length is 5λ/4 = 5×0.6666/4 = 0.83 m. This too is possible.
The fourth resonance will be obtained when the exposed length of the pipe is 7λ/4 = 7×0.6666/4 = 1.16 m. This is not possible since the length of the entire pipe is 1.1 m only.
So, the correct option is (c).
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