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

$A$ closed organ pipe has a fundamental frequency $n$. If its length is doubled and its radius is halved,its frequency nearly becomes

The fundamental frequency of a pipe is $100 \ Hz$ and the next two frequencies are $300 \ Hz$ and $500 \ Hz$. Then,the pipe is:

$A$ cylindrical tube $(L = 120 \, cm)$ is resonant with a tuning fork of frequency $330 \, Hz$. If it is being filled with water,then to get resonance again,the minimum length of the water column is ...... $cm$ $(v_{air} = 330 \, m/s)$.

The first resonance length of a resonance tube is $40\, cm$ and the second resonance length is $122\, cm$. The third resonance length of the tube will be... $cm$

Two closed organ pipes of length $100 \,cm$ and $101 \,cm$ produce $16$ beats in $20 \,s$. When each pipe is sounded in its fundamental mode,calculate the velocity of sound in $m/s$.