Question

A tube open at both ends has fundamental frequency 'n'. If one end of the tube is dipped in liquid to ${\left(\frac{1}{4}\right)}^{th}$ of its length, its new fundamental frequency $n_1$ is ____________.

• A. $n_1=\frac{n}{2}$
• B. $n_1=n$
• C. $n_1=2n$
• D. $n_1=4n$

Answer 1

The correct option is C $n_1=2n$

For the tube open at both ends, $\frac{{\lambda }_1}{2}=l$

For the tube closed at one end, $\frac{{\lambda }_2}{4}=\frac{l}{4}$

Speed of sound $=n{\lambda }_1=n_1{\lambda }_2$

$\Rightarrow n_1=\frac{n{\lambda }_1}{{\lambda }_2}=\frac{n\times 2l}{l}=2n$