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An electrically driven loudspeaker is placed near the open end of a resonance 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. Speed of sound in air = 340 m/s.

Abhishek Dogra , 6 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 6 Years ago

Dear student
 
f = v/4l
 
f = 320/ 4*50*10^-2 = 100 Hz
 
so the frequency of other harmonics willbe odd multiples of f = (2n+1) 100
 
according to the question harmonic mean should be between 20 Hz to 2 KHZ
 
n = { 0, 1, 2, …........, 9}

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