$A$ 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

In an open organ pipe,$v_3$ and $v_6$ are the $3^{\text{rd}}$ and $6^{\text{th}}$ harmonic frequencies,respectively. If $v_6 - v_3 = 2200 \text{ Hz}$,then the length of the pipe is . . . . . . mm. (Take the velocity of sound in air as $330 \text{ m/s}$.)

The fundamental frequency of an air column in pipe $A$ closed at one end is in unison with the second overtone of an air column in pipe $B$ open at both ends. The ratio of the length of the air column in pipe $A$ to that of the air column in pipe $B$ is:

$A$ pipe of $30 \text{ cm}$ long and open at both ends produces harmonics. Which harmonic mode of the pipe resonates with a $1.1 \text{ kHz}$ source? (Given: speed of sound in air $v = 330 \text{ ms}^{-1}$)

The fundamental frequencies of vibrations of an air column in a pipe open at both ends and in a pipe closed at one end are $n_1$ and $n_2$ respectively. If both pipes have the same length $L$,then:

$A$ source of frequency $340 \,Hz$ is kept above a vertical cylindrical tube closed at the lower end. The length of the tube is $120 \,cm$. Water is slowly poured in just enough to produce resonance. Then, the minimum height (velocity of sound $= 340 \,m/s$) of the water level in the tube for that resonance is (in $\,m$)