KCET · Physics · Waves and Sound
A cylindrical tube open at both the ends has a fundamental frequency of \(390 \mathrm{~Hz}\) in air. If \(\frac{1}{4}\) th of the tube is immersed vertically in water the fundamental frequency of air column is
- A \(260 \mathrm{~Hz}\)
- B \(130 \mathrm{~Hz}\)
- C \(390 \mathrm{~Hz}\)
- D \(520 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(A) \(260 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Fundamental frequency of cylindrical open tube
\(\mathrm{n}=\frac{\mathrm{v}}{2 \mathrm{~L}}=390 \mathrm{~Hz}\)
When it is immersed in water it becomes a closed tube of length \(\frac{3}{4}\) th of the initial length.
Therefore, its fundamental frequency is
\(\begin{aligned}\mathrm{n}^{\prime} &=\frac{\mathrm{V}}{4\left(\frac{3}{4} \mathrm{~L}\right)}=\frac{\mathrm{v}}{3 \mathrm{~L}}=\frac{2}{3}\left(\frac{\mathrm{V}}{2 \mathrm{~L}}\right) \\&=\frac{2}{3} \times 390 \mathrm{~Hz}=260 \mathrm{~Hz}\end{aligned}\)
\(\mathrm{n}=\frac{\mathrm{v}}{2 \mathrm{~L}}=390 \mathrm{~Hz}\)
When it is immersed in water it becomes a closed tube of length \(\frac{3}{4}\) th of the initial length.
Therefore, its fundamental frequency is
\(\begin{aligned}\mathrm{n}^{\prime} &=\frac{\mathrm{V}}{4\left(\frac{3}{4} \mathrm{~L}\right)}=\frac{\mathrm{v}}{3 \mathrm{~L}}=\frac{2}{3}\left(\frac{\mathrm{V}}{2 \mathrm{~L}}\right) \\&=\frac{2}{3} \times 390 \mathrm{~Hz}=260 \mathrm{~Hz}\end{aligned}\)
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