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The 4th overtone of a closed organ pipe is same as that of 3rd overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
A
8 : 9
B
9 : 7
C
9 : 8
D
7 : 9
Detailed Solution
For a closed pipe, $n_{cop}=(2M+1)^{th}$ harmonic $=(2\times4+1)\times\dfrac{V}{4\ell_c}=\dfrac{9V}{4\ell_c}$ (4th overtone). For an open pipe, $n_{oop}=(M+1)^{th}$ harmonic $=(3+1)\dfrac{V}{2\ell_o}=\dfrac{4V}{2\ell_o}$ (3rd overtone). Equating: $\dfrac{9V}{4\ell_c}=\dfrac{4V}{2\ell_o}\Rightarrow\dfrac{\ell_c}{\ell_o}=\dfrac{18}{16}=\dfrac{9}{8}$.