An open pipe immersed in water to half its length. The ratio of the fundamental frequency of the pipe before…
An open pipe immersed in water to half its length. The ratio of the fundamental frequency of the pipe before and after immersion in water is
$1: 2$
$1: 1$
$1: 3$
$1: 4$
Solution
Let $L_{o}$ be the length of air column in pipe. After immersion,
$L_{c}=\frac{L_{0}}{2}$
Fundamental frequency in open pipe, $f_{o}=\frac{v}{2 L_{0}}$
Fundamental frequency in closed pipe, $f_{c}=\frac{v}{L_{c}}$
$\therefore \quad \frac{f_{o}}{f_{c}}=\frac{\frac{v}{2 L_{o}}}{\frac{v}{4 L_{c}}}=2\left(\frac{L_{c}}{L_{o}}\right)=2\left(\frac{L_{o}}{2 L_{o}}\right)=1: 1$