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(C) The frequency of the $n^{	ext{th}}$ harmonic for a closed pipe is $f_c = \frac{n V_1}{4 \ell_1}$,where $n$ must be odd. For the $9^{	ext{th}}$ h

Vedclass · Accepted Answer

$\frac{20}{9} \ cm$

Vedclass · Answer

(C) The frequency of the $n^{	ext{th}}$ harmonic for a closed pipe is $f_c = \frac{n V_1}{4 \ell_1}$,where $n$ must be odd. For the $9^{	ext{th}}$ harmonic,$f_c = \frac{9 V_1}{4 \ell_1}$.The frequency of the $m^{	ext{th}}$ harmonic for an open pipe is $f_o = \frac{m V_2}{2 \ell_2}$. For the $4^{	ext{th}}$ harmonic,$f_o = \frac{4 V_2}{2 \ell_2} = \frac{2 V_2}{\ell_2}$.Given $f_c = f_o$,we have $\frac{9 V_1}{4 \ell_1} = \frac{2 V_2}{\ell_2}$.Since the speed of sound $V = \sqrt{\frac{B}{ho}}$,we substitute $V_1 = \sqrt{\frac{B}{ho_1}}$ and $V_2 = \sqrt{\frac{B}{ho_2}}$.$\frac{9}{4 \ell_1} \sqrt{\frac{B}{ho_1}} = \frac{2}{\ell_2} \sqrt{\frac{B}{ho_2}} \Rightarrow \frac{\ell_2}{\ell_1} = \frac{8}{9} \sqrt{\frac{ho_1}{ho_2}}$.Given $\ell_1 = 10 \ cm$ and $\frac{ho_1}{ho_2} = \frac{1}{16}$,we get $\sqrt{\frac{ho_1}{ho_2}} = \frac{1}{4}$.$\ell_2 = 10 	imes \frac{8}{9} 	imes \frac{1}{4} = \frac{20}{9} \ cm$.

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