MHT CET · Physics · Waves and Sound
The fundamental frequency of a closed pipe is \(400 \mathrm{~Hz}\). If \(1 / 3^{\text {rd }}\) pipe is filled with water, the frequency of \(2^{\text {nd }}\) harmonic of the pipe will be (Neglect end correction)
- A \(600 \mathrm{~Hz}\)
- B \(1800 \mathrm{~Hz}\)
- C \(1200 \mathrm{~Hz}\)
- D \(300 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(B) \(1800 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Fundamental frequency of closed pipe,
\(\begin{array}{l}
n=\frac{V}{4 L}=400 \\
V=400 \times 4 L
\end{array}\)
If \(\frac{1}{3}\) rd of pipe is filled with water, then remaining length of air column is \(L-\frac{L}{3}=\frac{2 L}{3}\). Now, fundamental frequency \(=\frac{v}{4\left(\frac{2 L}{3}\right)}=\frac{3 v}{8 L}\)
The second harmonic of the pipe \(=3 \times\) fundamental frequency
\(\begin{array}{l}
=3 \times\left(\frac{3 \mathrm{v}}{8 \mathrm{~L}}\right) \\
=3 \times \frac{3 \times 400 \times 4 \mathrm{~L}}{8 \mathrm{~L}} \\
=1800 \mathrm{~Hz}
\end{array}\)
\(\begin{array}{l}
n=\frac{V}{4 L}=400 \\
V=400 \times 4 L
\end{array}\)
If \(\frac{1}{3}\) rd of pipe is filled with water, then remaining length of air column is \(L-\frac{L}{3}=\frac{2 L}{3}\). Now, fundamental frequency \(=\frac{v}{4\left(\frac{2 L}{3}\right)}=\frac{3 v}{8 L}\)
The second harmonic of the pipe \(=3 \times\) fundamental frequency
\(\begin{array}{l}
=3 \times\left(\frac{3 \mathrm{v}}{8 \mathrm{~L}}\right) \\
=3 \times \frac{3 \times 400 \times 4 \mathrm{~L}}{8 \mathrm{~L}} \\
=1800 \mathrm{~Hz}
\end{array}\)
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