The fundamental frequency in an open organ pi | vedclass

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(A) The fundamental frequency of an open organ pipe of length $l'$ is given by $f_{open} = \frac{v}{2l'}$.The frequencies of a closed organ pipe of le

Vedclass · Accepted Answer

$13.33$

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(A) The fundamental frequency of an open organ pipe of length $l'$ is given by $f_{open} = \frac{v}{2l'}$.The frequencies of a closed organ pipe of length $l$ are given by $f_n = \frac{nv}{4l}$,where $n = 1, 3, 5, \dots$ (odd harmonics).The third harmonic of a closed organ pipe corresponds to $n = 3$,so $f_{closed, 3} = \frac{3v}{4l}$.According to the problem,$f_{open} = f_{closed, 3}$.Therefore,$\frac{v}{2l'} = \frac{3v}{4l}$.Simplifying for $l'$,we get $l' = \frac{4l}{6} = \frac{2l}{3}$.Given $l = 20 \ cm$,we have $l' = \frac{2 	imes 20}{3} = \frac{40}{3} \approx 13.33 \ cm$.

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is $20 \ cm$,the length of the open organ pipe is .... $cm$.

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