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Step-by-Step Solution
Step 1: Write the frequency of the third harmonic for a closed organ pipe
For a closed organ pipe of length $l$, the frequency of the $n^\text{th}$ harmonic is given by
$$
f_n = \frac{(2n - 1)\,v}{4\,l}.
$$
Here, $v$ is the speed of sound. For the third harmonic, $n = 2$ (because the harmonics for a closed pipe are
first harmonic: $n=1$ (1st),
second harmonic: $n=2$ (3rd),
and so on). Hence the frequency of the third harmonic is
$$
f_{\text{closed},3} = \frac{3\,v}{4\,l}.
$$
Step 2: Write the fundamental frequency for an open organ pipe
For an open organ pipe of length $l'$, the fundamental frequency (first harmonic) is
$$
f_{\text{open},1} = \frac{v}{2\,l'}.
$$
Step 3: Equate the two frequencies
The problem states that the fundamental frequency of the open pipe equals the third harmonic frequency of the closed pipe. Therefore,
$$
\frac{3\,v}{4\,l} \;=\; \frac{v}{2\,l'}.
$$
Step 4: Solve for the length of the open organ pipe
Canceling $v$ on both sides:
$$
\frac{3}{4\,l} = \frac{1}{2\,l'}
\quad\Longrightarrow\quad
l' = \frac{4\,l}{3 \cdot 2} = \frac{2l}{3}.
$$
Given $l = 20\text{ cm}$ (the length of the closed organ pipe),
$$
l' = \frac{2 \times 20}{3} = \frac{40}{3} \approx 13.3\text{ cm}.
$$
This rounds to about $13.2\text{ cm}$, which matches the correct answer.
Final Answer:
The length of the open organ pipe is approximately 13.2 cm.
