Question:
An open air pipe of length 80 cm has the second harmonic frequency equal to the fundamental frequency of a closed organ pipe. Find the length of the closed pipe.
An open air pipe of length 80 cm has the second harmonic frequency equal to the fundamental frequency of a closed organ pipe. Find the length of the closed pipe.
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For open pipe, \(f_n = n v/2L\); for closed pipe, \(f_n = (2n-1)v/4L\). Equate harmonics as needed.
Updated On: Jul 18, 2026
- 20 cm
- 40 cm
- 60 cm
- 10 cm
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The Correct Option is A
Solution and Explanation
Step 1: Recall fundamental frequencies.
For an open pipe of length \(L_o\), the \(n\)-th harmonic frequency:
\[ f_n = n \frac{v}{2 L_o} \]
For a closed pipe of length \(L_c\), the fundamental frequency:
\[ f_1 = \frac{v}{4 L_c} \]
Step 2: Identify given values.
Open pipe length \(L_o = 80 \, \text{cm}\). Second harmonic frequency of open pipe: \(f_2 = 2 \cdot \frac{v}{2 L_o} = \frac{v}{L_o}\).
Step 3: Equate frequencies.
\[ f_2 (\text{open}) = f_1 (\text{closed}) \implies \frac{v}{L_o} = \frac{v}{4 L_c} \implies L_c = \frac{L_o}{4} \]
Step 4: Substitute value.
\[ L_c = \frac{80}{4} = 20 \, \text{cm} \]
Step 5: Verify.
Second harmonic of open pipe \(f_2 = v/L_o\), fundamental of closed pipe \(f_1 = v/(4 L_c) = v/80\). Consistent.
Step 6: Final conclusion.
Hence, the length of the closed pipe is:
\[ \boxed{20 \, \text{cm}} \]
For an open pipe of length \(L_o\), the \(n\)-th harmonic frequency:
\[ f_n = n \frac{v}{2 L_o} \]
For a closed pipe of length \(L_c\), the fundamental frequency:
\[ f_1 = \frac{v}{4 L_c} \]
Step 2: Identify given values.
Open pipe length \(L_o = 80 \, \text{cm}\). Second harmonic frequency of open pipe: \(f_2 = 2 \cdot \frac{v}{2 L_o} = \frac{v}{L_o}\).
Step 3: Equate frequencies.
\[ f_2 (\text{open}) = f_1 (\text{closed}) \implies \frac{v}{L_o} = \frac{v}{4 L_c} \implies L_c = \frac{L_o}{4} \]
Step 4: Substitute value.
\[ L_c = \frac{80}{4} = 20 \, \text{cm} \]
Step 5: Verify.
Second harmonic of open pipe \(f_2 = v/L_o\), fundamental of closed pipe \(f_1 = v/(4 L_c) = v/80\). Consistent.
Step 6: Final conclusion.
Hence, the length of the closed pipe is:
\[ \boxed{20 \, \text{cm}} \]
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