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 = 320ms−1.
The resonance column apparatus is equivalent to a closed organ pipe.
Here I =80 cm =80×10−2,
mv =320 m/s
⇒n0=v4I=3204×50×10−2=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)