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 .

Options

A. `2`
B. `4`
C. `8`
D. `6`

Answer

D

Explanation

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.