An open tube is in resonance with string. If tube is dipped in water, so that 75 % of length of tube is inside water, then the ratio of the frequency (vo) of tube to string is

Question

An open tube is in resonance with string. If tube is dipped in water, so that 75 % of length of tube is inside water, then the ratio of the frequency (vo) of tube to string is (A)  vo  (B)  2vo  (C)  (2/3)vo  (D)  (3/2)vo

Tardigrade Admin · Accepted Answer

When open tube is dipped in water, it becomes a tube closed at one end.
Fundamental frequency for open tube is

v_{o} = \frac{v}{2 l}

Length available for resonance of closed tube is

0.25 l

.

∴ v_{c} = \frac{v}{4 ( 0.25 l )} = \frac{v}{2 l} \times 2 = 2 v_{o}

Q. An open tube is in resonance with string. If tube is dipped in water, so that $75%$ of length of tube is inside water, then the ratio of the frequency $(v_{o})$ of tube to string is

$v_{o}$

$2 v_{o}$

$\frac{2}{3} v_{o}$

$\frac{3}{2} v_{o}$

When open tube is dipped in water, it becomes a tube closed at one end.
Fundamental frequency for open tube is $v_{o} = \frac{v}{2 l}$
Length available for resonance of closed tube is $0.25 l$ .
$∴ v_{c} = \frac{v}{4 ( 0.25 l )} = \frac{v}{2 l} \times 2 = 2 v_{o}$

Q. An open tube is in resonance with string. If tube is dipped in water, so that 75% of length of tube is inside water, then the ratio of the frequency (vo​) of tube to string is

vo​

2vo​

32​vo​

23​vo​