Question

A cylindrical tube, open at both ends, has a fundamental frequency $f_{0}$ in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of the air column now is

• A. $3f_{0}/4$
• B. $f_{0}$
• C. $f_{0}/2$
• D. $2f_{0}$

Answer 1

Answer: (B)

$f_{0}$

Answer 2

$n_{\text{open}} = \frac{v}{2l_{\text{open}}}$

$n_{\text{closed}} = \frac{v}{4l_{\text{closed}}} = \frac{v}{4l_{\text{open}}/2} = \frac{v}{2l_{\text{open}}}$

$\left( As\mspace{6mu} l_{closed} = \frac{l_{open}}{2} \right)$, i.e. frequency remains unchanged.