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

Options

A. 4
B. 5
C. 7
D. 6

Answer

D

Explanation

Fundamental frequency of the closed organ pipe is `upsilon=(v)/(4L)`
Here, `v=340ms^(-1),L=85cm=0.85cm`
`therefore upsilon=(340ms^(-1))/(4xx0.85m)=100Hz`
The natural frequencies of the closed organ pipe will be
`upsilon_(n)=(2n-1)upsilon=upsilon,3upsilon,5upsilon,7upsilon,9upsilon,11upsilon,13upsilon,...`
`=100 Hz,300 Hz,500 Hz,700 Hz, 900 Hz, 1100 Hz, 1300 Hz,...` and so on.
Thus, the natural frequencies lies below the 1250 Hz is 6.