A cylindrical resonance tube, open at both ends, has a fundamental frequency $f$ in air. If half of the…
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 of the air column will be
$\frac{3f}{2}$
$2f$
$f$
$\frac{f}{2}$
Solution
Frequency, $f = \frac{v}{2l}$ and $f' = \frac{v}{4(l / 2)} = \frac{v}{2l} = f$
when half of the length is dipped in water, it will become closed pipe.