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Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20 cm. The speed of sound in air is 340 m/s. |
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Answer» The resonant frequencies in an open organ pipe is given as, νₙ = nV/2L, where V = 340 m/s, L = 20 cm =0.20 m →νₙ = n*(340/2*0.20) =n*3400/4 =n*850 Hz. For the fundamental frequency, n = 1, ν₀ = 1*850 Hz =850 Hz For the first overtone frequency, n = 2, ν₁ = 2*850 Hz = 1700 Hz For the second overtone frequency, n = 3, ν₂ = 3*850 Hz =2550 Hz |
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