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Step-by-Step Solution
Step 1: Identify the frequencies for closed and open organ pipes
• The fundamental frequency of a closed organ pipe of length $l_c$ is given by
$$ f_\text{closed} = \frac{v}{4\,l_c}, $$
where $v$ is the speed of sound in air.
• The fundamental frequency of an open organ pipe of length $l_0$ is
$$ f_\text{open} = \frac{v}{2\,l_0}. $$
• The second overtone in an open organ pipe corresponds to the third harmonic, which has a frequency
$$ f_{\text{2nd overtone}} = \frac{3v}{2\,l_0}. $$
Step 2: Set up the equality between the frequencies
We are told that the fundamental frequency of the closed pipe equals the second overtone of the open pipe. Mathematically:
$$ \frac{v}{4\,l_c} = \frac{3v}{2\,l_0}. $$
Step 3: Solve for the length of the open organ pipe
Canceling $v$ from both sides, we get:
$$ \frac{1}{4\,l_c} = \frac{3}{2\,l_0}
\quad \Longrightarrow \quad \frac{1}{l_c} = \frac{3 \times 2}{2 \times l_0}
\quad \Longrightarrow \quad \frac{1}{l_c} = \frac{3}{l_0}
\quad \Longrightarrow \quad l_0 = 3 \times l_c \times 2 = 6\,l_c. $$
Given $l_c = 20\text{ cm}$, we substitute:
$$ l_0 = 6 \times 20\text{ cm} = 120\text{ cm}. $$
Step 4: Conclude the answer
Hence, the length of the organ pipe open at both ends is $120\text{ cm}$.
