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Question

An electronically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of air column in the tube is 80 cm. The frequency of the loudspeaker can be varied between 20 Hz and 2 kHz. Find the frequencies at which the column will resonate. Speed of sound in air = 320ms1.


Solution

The resonance column apparatus is equivalent to a closed organ pipe.

Here I =80 cm =80×102,

mv =320 m/s

n0=v4I=3204×50×102=100Hz

So the frequency of the other harmonies are odd multiple of n_0

=(2n+1)100 Hz

According to the question, the harmonic should be between 20 Hz and 2 kHz

So, n=(0,1,2,3,4,5,...9)

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