Question:
The fundamental frequency of a closed pipe is \(400\) Hz. If \(\frac{1}{3}^{rd}\) pipe is filled with water then the frequency of \(3^{rd}\) harmonic of the pipe will be (Neglect end correction)
The fundamental frequency of a closed pipe is \(400\) Hz. If \(\frac{1}{3}^{rd}\) pipe is filled with water then the frequency of \(3^{rd}\) harmonic of the pipe will be (Neglect end correction)
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Filling one third with water shortens the air column to two thirds, which increases the fundamental frequency.
Updated On: Nov 12, 2026
- \(1200\) Hz
- \(1800\) Hz
- \(600\) Hz
- \(300\) Hz
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
A closed pipe of length \(L\) has fundamental \(f_0=\frac{v}{4L}=400\) Hz, and its harmonics are odd multiples of \(f_0\).
Step 2: New Air Column
Water fills \(\frac{1}{3}\) of the pipe, so the air column is \(L'=\frac{2L}{3}\).
Step 3: New Fundamental
\(f_0'=\frac{v}{4L'}=\frac{3}{2}\cdot\frac{v}{4L}=1.5\times 400=600\) Hz.
Step 4: Third Harmonic
\(3f_0'=3\times 600=1800\) Hz, option (B).
Final Answer:
The third harmonic frequency is 1800 Hz. \[ \boxed{\text{(B) }1800\ \text{Hz}} \]
A closed pipe of length \(L\) has fundamental \(f_0=\frac{v}{4L}=400\) Hz, and its harmonics are odd multiples of \(f_0\).
Step 2: New Air Column
Water fills \(\frac{1}{3}\) of the pipe, so the air column is \(L'=\frac{2L}{3}\).
Step 3: New Fundamental
\(f_0'=\frac{v}{4L'}=\frac{3}{2}\cdot\frac{v}{4L}=1.5\times 400=600\) Hz.
Step 4: Third Harmonic
\(3f_0'=3\times 600=1800\) Hz, option (B).
Final Answer:
The third harmonic frequency is 1800 Hz. \[ \boxed{\text{(B) }1800\ \text{Hz}} \]
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