A closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their length are in the ratio
Medium
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$f_{e}=\frac{(2 m+1) V}{4 L_{c}}$
${f_o} = \frac{{(m + 1)V}}{{2{L_o}}}$
For ${\rm{m}} = 1$ ( ${{1^{{\rm{st }}}}}$ overtone)
$\mathrm{f}_{\mathrm{c}}=\mathrm{f}_{\mathrm{o}}$
$\frac{{3{\text{V}}}}{{4{{\text{L}}_c}}} = \frac{{2{\text{V}}}}{{2{{\text{L}}_o}}}\quad \Rightarrow \quad \boxed{\frac{{{{\text{L}}_c}}}{{{{\text{L}}_o}}} = \frac{3}{4}}$
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