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An organ pipe closed at one end has fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear is |
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Answer» Solution :In case of closed pipe, `f = (2n + 1) f _(1)` (Where `f _(1)=` fundamental frequency, `2n +1= ` order of harmonic, `n = 0,1,2,3,…=` order of overtone) Note : If we take `f =(2n -1) f _(1),`then `2n -1=` order of harmonic, `n = 1,2,3…., n-1 =` order of overtone) Here `f = 20000 HZ and f _(1) = 1500 Hz and so 20000 = (2n +1) xx 1500` `therefore 2n +1 = 13.33` `therefore 2n +1 =13 (because ` here (2n +1) is less than 15 and an odd integer) `therefore n =6` `6^(th)` overtone is the MAXIMUM order os overtone (or highest overtone) that can be heard in this case. (Because if we take n = 7 then (2n +1) `f _(1) = 15 xx 1500 =22500` which would be GREATER than 20000 Hz and so it would be wrong). |
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