Question

A cylindrical tube, open at both ends, has a fundamental frequency v. The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is
(a) v/4
(b) v/2
(c) v
(d) 2v.

Answer 1

c) υ

If v is the velocity of the wave and L is the length of the pipe,
then the fundamental frequency for an open organ pipe is
$\nu =\frac{v}{2L}$

For a closed organ pipe of length L' = L/2, the fundamental frequency is
$\nu =\frac{v}{4L\text{'}}=\frac{v\times 2}{4\times L}=\frac{v}{2L}=v$

(When the pipe is dipped in water, it behaves like a closed organ pipe that is half the length)