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 $ (v_{o}) $ of tube to string is

• A. $ v_{o} $
• B. $ 2v_{o} $
• C. $ \frac{2}{3}v_{o} $
• D. $ \frac{3}{2}v_{o} $

Answer 1

Answer: (B)

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 $0.25 \,l$.
$\therefore v_{c}=\frac{v}{4(0.25 l)}=\frac{v}{2 l} \times 2=2 v_{o}$