Question

A cylindrical tube open at both the ends has a fundamental frequency of $390\,Hz$ in air. If $1/4^{th}$ of the tube is immersed vertically in water the fundamental frequency of air column is

• A. 130 Hz
• B. 390 Hz
• C. 520 Hz
• D. 260 Hz

Answer 1

Answer: (D)

260 Hz

Answer 2

Fundamental frequency of cylindrical open tube
$n=\frac{v}{2 L}=390 \,Hz$
When it is immersed in water it becomes a closed tube of length $\frac{3}{4}$ of the initial length.
Therefore, its fundamental frequency is
$n'=\frac{v}{4\left(\frac{3}{4} L\right)}=\frac{v}{3 L}=\frac{2}{3}\left(\frac{v}{2 L}\right)$
$=\frac{2}{3} \times 390 \,Hz =260\, Hz$