Question

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

Answer 1

The resonance column acts as a closed organ pipe for which the resonant frequencies are given as 

νₙ = (2n+1)V/4L, where n = 0, 1, 2, 3, .... 

Here V = 320 m/s, L = 80 cm = 0.80 m 

→νₙ = (2n+1)*320/(4*0.80) Hz 

→νₙ = 100(2n+1) Hz 

Putting n = 0, ν = 100 Hz > 20 Hz, Acceptable. 

for n = 10, ν = 2100 Hz > 2kHz, Not acceptable. 

For n = 9, ν = 1900 Hz < 2kHz, acceptable. 

Hence the frequencies at which the column will resonate  

= 100(2n+1) Hz where n = 0, 1, 2, 3,....., 9