In fundamental mode,the time required for the sound wave to reach up to the closed end of a pipe filled with air is $t$ seconds. The frequency of vibration of the air column is

$A$ tuning fork of frequency $340 \ Hz$ is vibrated just above a tube of $120 \ cm$ height. Water is slowly poured in the tube. What is the minimum height of water necessary for resonance (in $cm$)?

In an organ pipe whose one end is at $x = 0$,the pressure is expressed by $p = p_0 \cos \frac{3\pi x}{2} \sin 300\pi t$,where $x$ is in meters and $t$ is in seconds. The organ pipe can be:

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed organ pipe $L$ metre long. The length of the open pipe will be:

$A$ closed organ pipe of length $L$ and an open organ pipe contain gases of densities $\rho_1$ and $\rho_2$ respectively. The compressibility of the gases is equal in both the pipes. If the frequencies of their first overtones are the same,then the length of the open organ pipe is:

The length of an open organ pipe is $L$ and its fundamental frequency is $f$. If it is now immersed into water up to half of its length,what will be the new fundamental frequency of the organ pipe?