The fundamental frequency of a closed organ pipe of length $20\; cm$ is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is ...... $cm$.

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

$A$ closed organ pipe $150 \ cm$ long gives $7$ beats per second with an open organ pipe of length $350 \ cm$,both vibrating in fundamental mode. The velocity of sound is . . . . . . $m/s$.

Tube $A$ has both ends open while tube $B$ has one end closed,otherwise they are identical. The ratio of the fundamental frequency of tube $A$ to tube $B$ is

In an organ pipe closed at one end,the sum of the frequencies of the first three overtones is $3930 \ Hz$. The frequency of the fundamental mode of the organ pipe is: (in $Hz$)

The fundamental frequency of a closed pipe is $300 \ Hz$. What will be the frequency of its second overtone (in $Hz$)?