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(B) The frequency of the $m^{th}$ overtone for a closed organ pipe is given by $f_{c} = \frac{(2m+1)V}{4L_{c}}$,where $m=1$ for the first overtone,so

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$3:4$

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(B) The frequency of the $m^{th}$ overtone for a closed organ pipe is given by $f_{c} = \frac{(2m+1)V}{4L_{c}}$,where $m=1$ for the first overtone,so $f_{c} = \frac{3V}{4L_{c}}$.The frequency of the $m^{th}$ overtone for an open organ pipe is given by $f_{o} = \frac{(m+1)V}{2L_{o}}$,where $m=1$ for the first overtone,so $f_{o} = \frac{2V}{2L_{o}} = \frac{V}{L_{o}}$.Given that the frequencies are identical,$f_{c} = f_{o}$:$\frac{3V}{4L_{c}} = \frac{V}{L_{o}}$Rearranging the terms to find the ratio of lengths $L_{c}:L_{o}$:$\frac{L_{c}}{L_{o}} = \frac{3}{4}$Thus,the ratio of their lengths is $3:4$.

$A$ closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio:

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