The fundamental frequency of a closed organ pipe of a length $20 ext{ cm}$ is equal to the second overtone o

Question

The fundamental frequency of a closed organ pipe of a length $20 	ext{ cm}$ is equal to the second overtone of an organ pipe open at both ends. The length of the organ pipe open at both ends will be:

Accepted Answer

Let the length of the closed organ pipe be $L_c = 20 	ext{ cm}$.
The fundamental frequency of a closed organ pipe is given by $f_c = \frac{v}{4L_c}$.
Let the length of the open organ pipe be $L_o$.
The second overtone of an open organ pipe corresponds to the third harmonic. Its frequency is given by $f_o = \frac{3v}{2L_o}$.
According to the question, $f_c = f_o$:
$$\frac{v}{4L_c} = \frac{3v}{2L_o}$$
Rearranging to solve for $L_o$:
$$L_o = \frac{3v 	imes 4L_c}{2v} = 6L_c$$
Given $L_c = 20 	ext{ cm}$:
$$L_o = 6 	imes 20 	ext{ cm} = 120 	ext{ cm}$$

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