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Find out the frequencies of the fundamental and its nearest harmonic emitted by a 1 m long closed organ pipe. Given, velocity of sound in air ` 332 m * s^(-1)` .

Text Solution

Verified by Experts

Fundamental frequency, ` n_(0) = (V)/(4l) = (332)/(4 xx 1) = 83 ` Hz
A closed organ pipe can produce the odd harmonics only . So the frequency of the next harmonic,
` n_(1) = 3 n_(0) = 3 xx 83 = 249` Hz .
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