`AB` is a cylinder of length `1m` fitted with a thin flexible diaphragm `C` at the middle and other thin flexible diaphragms `A` and `B` at the ends. The portions `AC` and `BC` contain hydrogen and oxygen gases respectively. The diaphragms `A` and `B` are set into vibrations of same frequency. What is the minimum frequency of these vibrations for which diaphragms `C` is a node? (Under the conditions of experiment `v_(H_(2) = 1100m//s`, `v_(0_(2)` = `300m//s`).

Explanation

As diaphram C is a node. A and B will be antinode (as in a organ pipe either both ends are antinode or one end node and he other antinode) i.e., each part will behave as closed end organ pipe so that
`f_(H)=(V_(H))/(4L_(H))=(1100)/(4xx0.5)=550 Hz` And `f_(0)=(V_(0))/(4L_(0))=(300)/(4xx0.5)=150 Hz`
As the two fundamental frequencies are different, the system will vibrate with a common frequency f such that `f=n_(H)f_(H)=n_(0)f_(0)` i.e. `(n_(H))/(n_(0))=(f_(0))/(F_(H))=(150)/(550)=(3)/(11)`
i.e., the third harmonic of hydrogen and 11th harmonic of oxygen or 6th harmonic of hydrogen and 22nd harmonic of oxygen will have same frequency. So the minimum common frequency
` f=3xx550 or 11xx150=1650 Hz`