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  4. An open resonating tube has fundamental frequency of n . When half of its length is dipped into water then its fundamental frequency will be:

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Q. An open resonating tube has fundamental frequency of $n .$ When half of its length is dipped into water then its fundamental frequency will be:

BHUBHU 2002

Solution:

When half of the tube is dipped in water it behaves as a closed tube.
For on open pipe of length $l$ antinodes are formed at the two ends and node in between. Hence, fundamental frequency.
image
$n=\frac{v}{2 l}$
Fundamental frequency of closed pipe of length $\frac{l}{2}$,
forming antinode at open end and node at closed end is
$n'=\frac{v}{4\left(\frac{l}{2}\right)}=\frac{v}{2 l}=n=$ frequency of open pipe.