Two narrow cylindrical pipes `A` and `B` have the same length. Pipe `A` is open at both ends and is filled with a monoatomic gas of molar mass `M_(A)`. Pipe `B` is open at one end and closed at the other end and is filled with a diatomic gas of molar mass `M_(B)`. Both gases are at the same temperature.
(a) If the frequency of the second harmonic of the fundamental mode in pipe `A` is equal to the frequency of the third harmonic of the fundamental mode in pipe `B`, determine the value of `(m_(A))/(M_(B))`.
(b) Now, the open end of pipe `B` is also closed (so that the pipe is closed at both ends.) Find the ratio of the fundamental frequency in pipe `A` to that in pipe `B`.

Explanation

a. If `L` is the length of each pipe `A and B` , then fundamental frequency of pipe `A` ( open at both ends )
`n_(A) = (v_(A))/( 2L)` (i)
Fundamental frequency of pipe `B` ( closed at one end)
`n_(B) = (v_(B))/( 4L)` (ii)
Given `2 n_(A) = 3 n_(B)`
`2((v_(A))/(2L)) = 3((v_(B))/( 4L))`
`rArr (v_(A))/(v_(B)) = (3)/(4)`
But ` v_(A) = sqrt((gamma_(A) RT_(A))/( M_(A))) , v_(B) =sqrt((gamma_(B)RT_(B))/(M_(B)))`
Given `T_(A) = T_(B)`
For monoatomic get
`gamma _(A) = (5)/(3)`
For diatomic get
`gamma_(B) = (7)/(5)`
`:. (v_(A))/(v_(B)) = sqrt({(gamma_(A))/(gamma_(B))) (M_(B))/(M_(A))}`
`sqrt(((5//3)) /((7//5)) (M_(B))/(M_(A)))`
` = sqrt((25)/(21) (M_(B))/(M_(A)))`
(i) From Eqs.(iii) and (iv)
`sqrt((25)/(21) (M_(B))/(M_(A))) = (3)/(4)`
(ii) `(M_(A))/(M_(B)) = ((4)/(3))^(2) xx (25)/(21) = (400)/(189)`
b. When pipe `B` is closed at both ends , fundamental frequency of pipe `B` becomes
`n_(B) = (v_(B))/( 2L)` (v)
Using Eqs. (i) ,(iii) and (v) , we get
iii. `(n_(A))/(n_(B)) = (v_(A))/(v_(B)) = (3)/(4)`