The number of possible natural oscillartions of air column in a pipe closed at one end of length 85 cm whose ferquences lie below 1250 Hz are (velocity of sound = 340 ms^(-1))`

Options

A. `3`
B. `7`
C. `6`
D. `9`

Answer

C

Explanation

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`.