Question:
An open organ pipe and a closed organ pipe have the frequency of their second overtone identical. The ratio of length of closed pipe to that of open pipe is (Neglect end correction)
An open organ pipe and a closed organ pipe have the frequency of their second overtone identical. The ratio of length of closed pipe to that of open pipe is (Neglect end correction)
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Write the second overtone frequency for each pipe and equate them.
Updated On: Nov 13, 2026
\(2:3\)
\(5:6\)
\(3:4\)
\(6:7\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Open pipe: frequencies are \(f=\dfrac{nv}{2L}\) and its second overtone is the 3rd harmonic.
Closed pipe: frequencies are odd harmonics \(\dfrac{(2n-1)v}{4L}\) and its second overtone is the 5th harmonic.
Step 2: Frequencies
Open pipe: \(f_o=\dfrac{3v}{2L_o}\)
Closed pipe: \(f_c=\dfrac{5v}{4L_c}\)
Step 3: Calculation
Equate: \(\dfrac{3}{2L_o}=\dfrac{5}{4L_c}\)
\(\dfrac{L_c}{L_o}=\dfrac{5\times 2}{4\times 3}=\dfrac{5}{6}\).
Step 4: Compare the options
The ratio \(5:6\) is option (B).
Final Answer:
The ratio of closed pipe length to open pipe length is \(5:6\). \[ \boxed{\text{(B) }5:6} \]
Open pipe: frequencies are \(f=\dfrac{nv}{2L}\) and its second overtone is the 3rd harmonic.
Closed pipe: frequencies are odd harmonics \(\dfrac{(2n-1)v}{4L}\) and its second overtone is the 5th harmonic.
Step 2: Frequencies
Open pipe: \(f_o=\dfrac{3v}{2L_o}\)
Closed pipe: \(f_c=\dfrac{5v}{4L_c}\)
Step 3: Calculation
Equate: \(\dfrac{3}{2L_o}=\dfrac{5}{4L_c}\)
\(\dfrac{L_c}{L_o}=\dfrac{5\times 2}{4\times 3}=\dfrac{5}{6}\).
Step 4: Compare the options
The ratio \(5:6\) is option (B).
Final Answer:
The ratio of closed pipe length to open pipe length is \(5:6\). \[ \boxed{\text{(B) }5:6} \]
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