Question

A cylindrical resonance tube open at both ends has a fundamental frequency f in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be

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

Answer 1

The correct option is B $f$

Fundamental frequency of open tube ${\nu }_o=\frac{v}{2L}=f$ -----(1)

When half length of tube is dipped in water it acts as $\frac{L}{2}$ length of closed pipe.

So, it's fundamental frequency will be ${\nu }_c=\frac{v}{4\left(\frac{L}{2}\right)}=\frac{v}{2L}=f$ (From (1) )