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(A) The fundamental frequency of an open cylindrical tube of length $L$ is given by $n = \frac{v}{2L} = 390 \,Hz$.When $\frac{1}{4}$th of the tube is

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$260$

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(A) The fundamental frequency of an open cylindrical tube of length $L$ is given by $n = \frac{v}{2L} = 390 \,Hz$.When $\frac{1}{4}$th of the tube is immersed in water, the tube acts as a closed organ pipe (closed at one end) with a new length $L' = L - \frac{1}{4}L = \frac{3}{4}L$.The fundamental frequency of a closed organ pipe is given by $n' = \frac{v}{4L'}$.Substituting $L' = \frac{3}{4}L$, we get $n' = \frac{v}{4(\frac{3}{4}L)} = \frac{v}{3L}$.We can rewrite this as $n' = \frac{2}{3} 	imes (\frac{v}{2L})$.Since $\frac{v}{2L} = 390 \,Hz$, we have $n' = \frac{2}{3} 	imes 390 \,Hz = 260 \,Hz$.

$A$ cylindrical tube open at both ends has a fundamental frequency of $390 \,Hz$ in air. If $\frac{1}{4}$th of the tube is immersed vertically in water, what is the fundamental frequency of the air column (in $\,Hz$)?

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