Question

A cylindrical tube open both ends has a fundamental frequency n in air. The tube is dipped vertically in water so that one - fourth of it is immersed in water. The fundamental frequency of air column is

• A. $3n$
• B. $2/3n$
• C. $1/3n$
• D. $n$

Answer 1

Answer: (B)

$2/3n$

Open fundalental frequency ${\nu }_o=\frac{v}{2L}=n$ -----(1)

When tube is dipped so, that ${\frac{1}{4}}^{th}$ of it length is immersed it acts as closed pipe of length $\frac{3}{4}\times L$

So, fundamental frequency of this closed pipe is

${\nu }_c=\frac{v}{4\left(\frac{3}{4}L\right)}=\frac{v}{3L}$

So, from (1)

${\nu }_c=\frac{v}{3L}=\frac{2v}{3\times 2L}=\frac{2}{3}n$.