Question

A cylindrical tube, open at both ends has a fundamental frequency $ f $ in air. The tube dipped vertically in water, so that half of it is in water, the fundamental frequency of the air colon is now:

• A. $ 2\,f $
• B. $ f $
• C. $ \frac{3f}{4} $
• D. $ \frac{f}{2} $

Answer 1

Answer: (B)

When tube is dipped in water its acts like a closed end tube.
Frequency of fundamental tone in a cylindrical pipe open at both end's is

$f=\frac{v}{2 l}\,\,\,...(1)$
where $v$ is velocity of sound and $l$ is length of pipe.
When half the length is dipped, it acts like a closed end tube,
with fundamental frequency
$f'=\frac{v}{4 l'}$
Given, $l'=\frac{l}{2}$
$\therefore f'=\frac{v}{4\left(\frac{l}{2}\right)}\,\,\,\,...(2)$
Comparing Eqs. (1) and (2), we get
$f'=f$
Therefore, the fundamental frequency of air column is same as that for open one.