Stationary waves

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Question

nth harmonic of a closed organ pipe is equal to mth harmonic of a open pipe. First overtone frequency of the closed organ pipe is also equal to first overtone frequency of the open organ pipe. Find the value of n, if m = 6.

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Solution

$n (\frac{V}{4 L_{c}}) = m (\frac{V}{2 L_{0}})$
Also, $3 (\frac{V}{4 L_{c}}) = 2 (\frac{V}{2 L_{0}})$

From eq. (2) $\frac{L_{c}}{L_{0}} = \frac{3}{4}$

From eq. (1) $\frac{n}{m} = 2 (\frac{L_{c}}{L_{0}}) = \frac{6}{4} = \frac{3}{2} = \frac{9}{6}$

$n = 9 i f m = 6$

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