When diaphragms `A` and `B` are set in oscillations , antinodes are formed at `A` and `B` while a node is formed at `C` ( given)
Also `AB = 1.0 m`
and ` AC = CB = l (say) = 1//2 = 0.5 m`
The portions `AC and BC` behave as closed pipes .
In a closed pipe , the modes of vibrations are given by
`l = (lambda)/(4) , ( 3lambda)/(4), ( 5 lambda)/(4) , ....`
i.e., ` l = ( 2 r + 1) (lambda)/( 4) , r = 0 , 1 , 2 , 3 ,...`
or ` lambda = ( 4l)/(2 r + 1) r = 0 , 1 , 2, 3 ,....`
For hydrogen ` lambda_(1) = 4l//( 2 r_(1) + 1)`
For oxygen ` lambda_(2) = 4 l//( 2r_(2) + 1)`
In both gases the frequency is same
As ` n_(1) = n_(2)`
or `(v_(1))/( lambda_(1)) = (v_(2))/(lambda_(2))`
`(v_(1))/(v_(2)) = (lambda_(1))/( lambda) = (l_(1))/(l_(2)) xx (( 2 r_(2) + 1))/(( 2 r_(1) + 1))`
As `l_(1) = l_(2) = 0.5 m`
i.e., ` (v_(1))/( v_(2)) = ( 2 r_(2) + 1) /( 2 r_(1) + 1)`
` = (1100)/(300) = ( 2r_(2) + 1)/( 2 r_(1) + 1)`
i.e., `( 2 r_(1) + 1)/( 2 r_(2) + 1) = (3)/(11)`
For minimum frequency , the integer `r_(1) and r_(2)` should be least . Therefore by the inspection
` r_(1) = 1 and r_(2) = 5`
Therefore , the frequency of oscillations is given by
`n_(min) = (v_(1))/( lambda_(1)) = ( 2 r_(1) + 1) ( v_(1))/( 4 l)`
`= ( 2xx 1 + 1) xx ( 1100)/(4 xx 0.5) = ( 3 xx 1100)/(2) = 1650 Hz`
i.e., `( 2 r_(1) + 1)/( 2 r_(2) + 1) = (3)/(11)`
For minimum frequency , the integers `r_(1) and r_(2)` should be least. Therefore by inspection
`r_(1) = 1and r_(2) = 5`
Therefore , the frequency of oscillations is given by
`n_(min) = (v_(1))/( lambda_(1)) = ( 2 r_(1) + 1) v_(1)/( 4 l)`
` = ( 2 xx 1 + 1) xx ( 1100)/( 4 xx 0.5) = ( 3 xx 1100)/(2) = 1650 Hz`
