Find the number of possible natural oscillations of air column in a pipe whose frequencies lie below f_(0) = 1250 Hz. The length of the pipe is l = 85 cm. The velocity of sound is v = 340 m//s. Consider two cases the pipe is opened from both ends.

Find the number of possible natural oscillations of air column in a pi

1.	Find the number of possible natural oscillations of air column in a pipe whose frequencies lie below f_(0) = 1250 Hz. The length of the pipe is l = 85 cm. The velocity of sound is v = 340 m//s. Consider two cases the pipe is opened from both ends.
Answer» `3` `7` `6` `9` Solution :Pipe opened from both ends Fundamental frequency `= (V)/( 2 l) = (340)/(2 xx 85) xx 100 = 200 Hz` Frequency of the other natural modes of oscillational are : `2 nd` harmonic frequency ` = 2 xx 200 = 400 Hz ` `3 rd` harmonic frequency ` = 3 xx 200 = 600 Hz` `4 TH` harmonic frequency ` = 4 xx 200 = 800 Hz` `5 th` harmonic frequency ` = 6 xx 200 = 1200 Hz` `7 th` harmonic frequency`= 7 xx 200 = 1400 Hz` You have to count only those harmonics whose frequencies are below `1250 Hz`. All the harmonics till `6 th` harmonic are possible , and obviously they are six in number . Second Method Fundamental frequency ` = 200 Hz` ` = n xx` fundamental frequency `= n xx 200``[ n is 1 , 2, ....]` Maximum value of `n = [ 12.50 //200] = 6 ([x]` REPRESENTS greatest less than or equal to `x` ). Now `n` is ALSO equal to the number of harmonic for which frequency is being calculated , highest harmonic possible ` = 6 th`. A all harmonics are possible in case of open tube , harmonics possible are `1 st , 2 nd , 3 rd , 4 th, 5 th , 6 th. ` Number of harmonics possible in this case `= 6`.