First overtone frequency of a closed organ pipe is equal to the first overtone frequency of an open organ…

First overtone frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. Further, $n$th harmonic of closed organ pipe is also equal to the $m$th harmonic of open pipe, where the respective values of $n$ and $m$ are
  1. (a) 5, 4
  2. (b) 7, 5
  3. (c) 9, 6
  4. (d) 7, 3

Solution

Given, $3 \left(\frac{v}{4l_c}\right) = 2 \left(\frac{v}{2l_o}\right)$ or $\frac{l_c}{l_o} = \frac{3}{4}$ Now, $n \left(\frac{v}{4l_c}\right) = m \left(\frac{v}{2l_o}\right)$ or $\frac{n}{m} = 2 \frac{l_c}{l_o} = \frac{3}{2} = \frac{9}{6}$ Thus, ratio $n/m$ should be $3/2$ but $n$ is only odd, while $m$ may be even or odd.

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