An air column in a tube 32 ~cm long,closed at | vedclass

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(C) The fundamental frequency of a closed organ pipe is given by $n_1 = \frac{v}{4l_1}$.Given $l_1 = 32 ~cm = 0.32 ~m$,so $n_1 = \frac{v}{4 	imes 0.3

Vedclass · Accepted Answer

$264 ~Hz, 256 ~Hz$

Vedclass · Answer

(C) The fundamental frequency of a closed organ pipe is given by $n_1 = \frac{v}{4l_1}$.Given $l_1 = 32 ~cm = 0.32 ~m$,so $n_1 = \frac{v}{4 	imes 0.32} = \frac{v}{1.28}$.The fundamental frequency of an open organ pipe is given by $n_2 = \frac{v}{2l_2}$.Given $l_2 = 66 ~cm = 0.66 ~m$,so $n_2 = \frac{v}{2 	imes 0.66} = \frac{v}{1.32}$.The beat frequency is $|n_1 - n_2| = 8 ~Hz$.Thus,$\frac{v}{1.28} - \frac{v}{1.32} = 8$.$\frac{1.32v - 1.28v}{1.28 	imes 1.32} = 8$.$0.04v = 8 	imes 1.28 	imes 1.32$.$v = \frac{8 	imes 1.6896}{0.04} = 200 	imes 1.6896 = 337.92 ~m/s$.Now,calculate the frequencies:$n_1 = \frac{337.92}{1.28} = 264 ~Hz$.$n_2 = \frac{337.92}{1.32} = 256 ~Hz$.

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