A closed organ pipe of length L and an open o | vedclass

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Question

(A) For a closed organ pipe,the frequency of the first overtone is given by $f_{c} = \frac{3v_{1}}{4L}$,where $v_{1} = \sqrt{\frac{B}{\rho_{1}}}$ and

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$4$

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(A) For a closed organ pipe,the frequency of the first overtone is given by $f_{c} = \frac{3v_{1}}{4L}$,where $v_{1} = \sqrt{\frac{B}{\rho_{1}}}$ and $B$ is the bulk modulus.For an open organ pipe,the frequency of the first overtone is given by $f_{o} = \frac{2v_{2}}{2L'} = \frac{v_{2}}{L'}$,where $v_{2} = \sqrt{\frac{B}{\rho_{2}}}$.Given that the frequencies are equal,$f_{c} = f_{o}$,so $\frac{3}{4L} \sqrt{\frac{B}{\rho_{1}}} = \frac{1}{L'} \sqrt{\frac{B}{\rho_{2}}}$.Rearranging for $L'$,we get $L' = \frac{4L}{3} \sqrt{\frac{\rho_{1}}{\rho_{2}}}$.Comparing this with the given expression $L' = \frac{x}{3} L \sqrt{\frac{\rho_{1}}{\rho_{2}}}$,we find that $x = 4$.

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