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Two successive resonant frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air =324 m/s. |
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Answer» The resonant frequencies in an open organ pipe is given as νₙ = nV/2L Let the given frequencies are for n and n+1 harmonics, thus nV/2L = 1944 and (n+1)V/2L = 2592 →nV/2L + V/2L =2592 →1944+V/2L = 2592 →V/2L = 2592 - 1944 = 648 →L = V/(2*648) = 324/1296 m = 0.25 m →L = 25 cm |
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