The air column in an organ pipe closed at one end is made to vibrate so that there are $2$ nodes and $2$ antinodes each. The mode of vibration is called:

If $v$ is the speed of sound in air,then what is the shortest length of a closed pipe that resonates at a frequency $n$?

In the fundamental mode, the time required for a sound wave to reach the closed end of a pipe filled with air is $t$ seconds. What is the frequency of vibration of the air column? ($\lambda$ = wavelength of the wave)

$A$ 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 $\,m$)

$A$ closed organ pipe and an open organ pipe are filled with two different gases having the same bulk modulus but different densities $\rho_1$ and $\rho_2$ respectively. The frequency of the $9^{\text{th}}$ harmonic of the closed pipe is identical to the $4^{\text{th}}$ harmonic of the open pipe. If the length of the closed pipe is $10 \ cm$ and the density ratio of the gases is $\rho_1 : \rho_2 = 1 : 16$,then the length of the open pipe is:

$A$ pipe $30 \ cm$ long is open at both ends. Which harmonic mode of the pipe is resonantly excited by a $1.1 \ kHz$ source? (Take speed of sound in air = $330 \ ms^{-1}$)