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

Frequency of the fundamental note produced by closed end organ pipe,
`v_(0)=(upsilon)/(4l)=(340ms^(-1))/(4xx0.85m)=100Hz`
Frequencies of other notes produced by pipe
`v_(n)=(2n_1)v_(0)` where`n=0,1,2,3,4,.....`
`=v_(0), 3v_(0), 5v_(0), 7v_(0), 9v_(0), 11v_(0), 13v_(0)`
`=100Hz, 300Hz, 500Hz, 700Hz, 900Hz, 1100Hz, 1300Hz`
Thus, the number of notes below the frequency `1250Hz` will be 6.