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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 closed from one end .A. `2`B. `4`C. `8`D. `6` |
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Answer» Correct Answer - D Pipe is closed from one end : An air column in a pipe closed from one end oscillates only odd harmonics [ `I st` harmonic ( fundamental mode) , `3 rd` harmonic ( `Ist` overtone) , `5 th` harmonic ( `2 nd` overtone) , `7 th` harmonic ( `3 rd` overtone) etc.] Fundamental frequency ` = ( V)/( 4l) = ( 340)/( 4 xx 85/100) = 100 Hz` Other modes of oscillation are ` 3 rd` harmonic frequency ` = 3 xx 100 = 300 Hz` `5 th` harmonic frequency `= 5 xx 100 = 500 Hz` `7 th` harmonic frequency ` = 7 xx 100 = 700 Hz` `9 th` harmonic frequency ` = 9 xx 100 = 900 Hz` `11 th` harmonic frequency ` = 11 xx 100 = 1100 Hz` `13 th` harmonic frequency ` = 13 xx 100 = 1300 Hz` Only those natural oscillations are to be counted whose frequencies lie below `f_(0) = 1250 Hz` , the harmonics till `11 th` harmonic are to be counted . Since , the number of piossible natural oscillations `= 1 ( i st "harmonic") + 1( 3 rd "harmonic") + 1( 5 th "harmonic") + 1( 7 th "harmonic") + 1( 9 th "harmonic") + 1 ( 11 th "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 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 , highest harmonic possible ` = 11 th` . Th epossible harmonics are `1 , 3 , 5 ,5 , 7 , 9 , 11` where are six in number. |
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