A cylindrical tube, open at both ends, has a fundamental frequency $f_{0}$ , in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of the air column now is

\frac{3 f_{0}}{4}

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f_{0}

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\frac{f_{0}}{2}

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3 f_{0}

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Solution

The correct option is B $f_{0}$
$f_{o p e n} = \frac{u}{2 L} a n d f_{c l o s e d} = \frac{u}{4 L / 2} = \frac{u}{2 L}$
Clearly $f_{o p e n} = f_{c l o s e d}$

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Similar questions

f_{o}

A cylindrical tube, open at both ends, has a fundamental frequency $f_{0}$ in air.

The tube is dipped vertically into water such that half of its length is inside

water. The fundamental frequency of the air column now is

[RPET 1999; RPMT 1998, 2000; J & K CET 2000;

KCET 2002; BHU 2002; BCECE 2003]

f_{0}

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