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

Question

An open resonating tube has fundamental frequency of n . When half of its length is dipped into water then its fundamental frequency will be: (A)  (3/2)n  (B) (n/2) (C) 2n (D)  n

Tardigrade Admin · Accepted Answer

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.

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 ( \frac{l}{2} )} = \frac{v}{2 l} = n =

frequency of open pipe.

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:

$\frac{3}{2} n$

$\frac{n}{2}$

$2 n$

$n$

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:

23​n

2n​

2n

n

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:

$\frac{3}{2} n$

$\frac{n}{2}$

$2 n$

$n$