The second overtone of an open organ pipe A a | vedclass

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(D) For an open organ pipe $A$,the frequency of the $n$-th overtone is given by $
u_{A,n} = (n+1) \frac{v}{2 l_A}$. The second overtone $(n=2)$ is $\

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Both $(B)$ and $(C)$

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(D) For an open organ pipe $A$,the frequency of the $n$-th overtone is given by $
u_{A,n} = (n+1) \frac{v}{2 l_A}$. The second overtone $(n=2)$ is $
u_{A,2} = \frac{3v}{2l_A}$.For a closed organ pipe $B$,the frequency of the $n$-th overtone is given by $
u_{B,n} = (2n+1) \frac{v}{4 l_B}$. The second overtone $(n=2)$ is $
u_{B,2} = \frac{5v}{4l_B}$.Given $
u_{A,2} = 
u_{B,2}$,we have $\frac{3v}{2l_A} = \frac{5v}{4l_B}$.Rearranging gives $\frac{l_B}{l_A} = \frac{5 	imes 2}{3 	imes 4} = \frac{10}{12} = \frac{5}{6}$. Thus,the ratio of lengths of $B$ to $A$ is $5:6$.Now,the first overtone of $A$ $(n=1)$ is $
u_{A,1} = \frac{2v}{2l_A} = \frac{v}{l_A}$.The first overtone of $B$ $(n=1)$ is $
u_{B,1} = \frac{3v}{4l_B}$.The ratio of these frequencies is $\frac{
u_{A,1}}{
u_{B,1}} = \frac{v}{l_A} 	imes \frac{4l_B}{3v} = \frac{4}{3} 	imes \frac{l_B}{l_A} = \frac{4}{3} 	imes \frac{5}{6} = \frac{20}{18} = \frac{10}{9}$.Both statements $(B)$ and $(C)$ are correct.

The second overtone of an open organ pipe $A$ and a closed pipe $B$ have the same frequency at a given temperature. It follows that the ratio of the

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