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Find the number of possible natural oscillations of air column in a pipe frequencies of which lie belowv_(0) = 1250 Hz. The length of the pipe is l = 85 cm. The velocity of sound is v = 340 m//s. Consider two cases i. the pipe is closed from the end , ii. the pipe is open from both ends. |
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Answer» Solution :First Method i. Pipe is closed from one end : An air column in a pipe closed from one end oscillates only odd harmonics[ 1st harmonic (fundamental mode ), 3rd harmonic ( 1st overtone ), 5th harmonic ( 2nd overtone ) , 7th harmonic ( 3rd overtone ) etc.] Fundamental frequency ` = (v)/( 4 l) = ( 340 )/( 4 xx (85)/(100)) = 100 Hz` Other modes of oscillation are : `3rd "harmonic frequency" = 3 xx 100 = 300 Hz` ` 5th "harmonic frequency"= 5 xx 100 = 500 Hz` `7th "harmonic frequency" = 7 xx 100 = 700 Hz` ` 9th "harmonic frequency" = 9 xx 100 = 900 Hz` `11th "harmonic frequency" = 11 xx 100 = 1100 Hz` `13th "harmonic frequency" = 13 xx 100 = 1300 Hz` Only those natural oscillations are to be counted frequencies of which lie below ` v_(0) = 1250 Hz`, the harmonics till 11th harmonic are to be counted. Since number of possible natural oscillations ` = 1 ( 1st "harmonic") + 1 ( 3rd "harmonic") + 1 ( 5th "harmonic" ) + 1 ( 7th "harmonic" ) + 1( 9th "harmonic" ) + 1 ( 11th "harmonic" ) = 6` second Method All the frequencies possible are integral MULTIPLES of fundamental frequency which is `100 Hz`. Using the fact that integer which is multiplied by fundamental frequency is the number of harmonic itself you get , highest harmonic predicted `= [ 12.50 // 100 ] ` where `[ x]` represents greatest integer less than or equal to ` x = [ 12.5 ] = 12`. Now for closed pipe , only odd harmonics are possible and the highest harmonic possible = 11 th . The possible harmonics are `1 , 3 , 5 , 7 , 9 , 11` which are six in number. ii. Pipe opened from the both ends : Fundamental frequency ` = (V)/( 2 l) = ( 340)/( 2 xx 85) xx 100 = 200 Hz` Frequency of the other natural modes of oscillational are : ` 2nd "harmonic frequency" =2 xx 200 = 400 Hz` `3rd "harmonic frequency" = 3 xx 200 = 600 Hz` ` 4TH "harmonic frequency" = 4 xx 200 = 800 Hz` ` 5th "harmonic frequency"= 5 xx 200 = 1000 Hz` `6th "harmonic frequency" = 6 xx 200 = 1200 Hz` ` 7th "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. Third Method Fundamental frequency ` = 200 Hz` Frequencies possible ` = n xx "fundamental frequency"` ` = n xx 200 ``[ n is 1 , 2, ...]` Maximum value of ` n = [ 12.50 // 200] = 6 ([x] "represents greater than or equal to" x)` Now ` n` is also equal to the number of harmonics for which frequency is being calculated , highest harmonic possible `= 6 th`. As all harmonics are possible in case of open tube , harmonics possible are `1st, 2nd , 3rd , 4th , 5th and 6th`. Number of harmonics possible in this case ` = 6`. |
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