The fundamental note produced by a closed organ pipe is of frequency $f$. The fundamental note produced by an open organ pipe of the same length will be of frequency .... $(f)$

An open pipe of length $33 \, cm$ resonates to a frequency of $1000 \, Hz$. The mode of vibration is: (velocity of sound $= 330 \, m/s$)

$A$ pipe closed at one end produces a fundamental note of $412 \, Hz$. It is cut into two pieces of equal length. The fundamental notes produced by the two pieces are:

$A$ glass tube $1.5 \ m$ long and open at both ends is immersed vertically in a water tank completely. $A$ tuning fork of $660 \ Hz$ is vibrated and kept at the upper end of the tube,and the tube is gradually raised out of the water. Taking the velocity of sound in air as $330 \ m/s$,find the total number of resonances heard before the tube comes out of the water.

If $L_1$ and $L_2$ are the lengths of the first and second resonating air columns in a resonance tube,then the wavelength of the note produced is

The length of an open organ pipe is $0.5\, m$. Calculate the fundamental frequency of the pipe,if the velocity of sound in air is $350\, m/s$.