Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air is `324 m s^-1`.

Answer

B, C

Explanation

Let the length of the resoN/Ating column will be =I
Here `v=320 m/s
Then the two successie resoN/Ance
frequencies are((n+1))/(4I) and (nv)/(4I)`
Here given `((n+1)v)/(4I)=2592`
`=lamda =(nv)/(4I)=2592`
`lamda=(nv)/(4I)=1944`
`rarr((n+1)v)/(4I)-(nv)/(4I)=2592-1944=548`
`rarr v/(4I)=548`
`rarr I=(320xx100)/(4x548)cm=25cm`