A pipe of length 85 cm is closed from one end.Find the number of possible natural oscillations of air columm in the pipe whose frequencies lie bnelow 1250 Hz.The velocity of sound in air is 340 `m//s`.

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.