The first overtone frequency of a closed orga | vedclass

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(C) Let $L'$ and $L$ be the lengths of the closed and open organ pipes,respectively.For a closed organ pipe,the frequency of the $n^{th}$ harmonic is

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$9, 6$

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(C) Let $L'$ and $L$ be the lengths of the closed and open organ pipes,respectively.For a closed organ pipe,the frequency of the $n^{th}$ harmonic is given by $
u_n' = \frac{n v}{4 L'}$,where $n$ must be an odd integer $(n = 1, 3, 5, \dots)$.The first overtone of a closed pipe corresponds to the $3^{rd}$ harmonic $(n=3)$,so $
u_{1st, closed} = \frac{3 v}{4 L'}$.For an open organ pipe,the frequency of the $m^{th}$ harmonic is given by $
u_m = \frac{m v}{2 L}$,where $m = 1, 2, 3, \dots$.The first overtone of an open pipe corresponds to the $2^{nd}$ harmonic $(m=2)$,so $
u_{1st, open} = \frac{2 v}{2 L} = \frac{v}{L}$.Given $
u_{1st, closed} = 
u_{1st, open}$,we have $\frac{3 v}{4 L'} = \frac{v}{L}$,which implies $\frac{L'}{L} = \frac{3}{4}$.Given the $n^{th}$ harmonic of the closed pipe equals the $m^{th}$ harmonic of the open pipe: $\frac{n v}{4 L'} = \frac{m v}{2 L}$.Substituting $\frac{L'}{L} = \frac{3}{4}$,we get $\frac{n}{4(3/4)} = \frac{m}{2} \implies \frac{n}{3} = \frac{m}{2} \implies 2n = 3m$.Testing the options: For $n=9$ and $m=6$,$2(9) = 18$ and $3(6) = 18$. Thus,$n=9$ and $m=6$ is the correct pair.

The first overtone frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. Further,the $n^{th}$ harmonic of the closed organ pipe is also equal to the $m^{th}$ harmonic of the open pipe,where $n$ and $m$ are:

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