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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)`).A. 4B. 5C. 7D. 6 |
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Answer» Correct Answer - D 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 6. |
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