A 1 ; m long (both ends open) organ pipe is k | vedclass

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(B) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma P}{\rho}}$.Assuming $\gamma$ and $P$ are the same for the gas and air,we have $\fr

Vedclass · Accepted Answer

$106$

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(B) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma P}{ho}}$.Assuming $\gamma$ and $P$ are the same for the gas and air,we have $\frac{v_{gas}}{v_{air}} = \sqrt{\frac{ho_{air}}{ho_{gas}}}$.Given $ho_{gas} = 2 ho_{air}$,we get $\frac{v_{gas}}{300} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus,$v_{gas} = \frac{300}{\sqrt{2}} = 150 \sqrt{2} \; m/s$.For an open organ pipe of length $L$,the fundamental frequency is $f_1 = \frac{v}{2L}$ and the second harmonic is $f_2 = \frac{2v}{2L} = \frac{v}{L}$.The frequency difference is $\Delta f = f_2 - f_1 = \frac{v}{2L}$.Substituting the values,$\Delta f = \frac{150 \sqrt{2}}{2(1)} = 75 \sqrt{2} \approx 75 	imes 1.414 = 106.05 \; Hz$.Rounding to the nearest integer,the answer is $106 \; Hz$.

$A$ $1 \; m$ long (both ends open) organ pipe is kept in a gas that has double the density of air at $STP$. Assuming the speed of sound in air at $STP$ is $300 \; m/s$,the frequency difference between the fundamental and second harmonic of this pipe is . . . . . . $Hz$.

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