A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 m s-1).
First (Fundamental); No
Length of the pipe, l = 20 cm = 0.2 m
Source frequency = nth normal mode of frequency, Vn = 430 Hz
Speed of sound, v = 340 m/s
In a closed pipe, the nth normal mode of frequency is given by the relation:
Vn = (2n -1) v / 4l ; n is an integer = 0, 1, 2, 3...
430 = (2n-1) 340 / (4x0.2)
2n -1 = (430 x 4 x 0.2) / 340 = 1.01
2n = 2.01
n ≈ 1
Hence, the first mode of vibration frequency is resonantly excited by the given source.
In a pipe open at both ends, the nth mode of vibration frequency is given by the relation:
Vn = nv / 2l
n = 2lVn / v
= 2 x 0.2 x 430 / 340 = 0.5
Since the number of the mode of vibration (n) has to be an integer, the given source does not produce a resonant vibration in an open pipe.
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