A cylinder of length `1m` is divided by a thin perfectly flexible diaphragm in the middle. It is closed by similar flexible diaphragams at the ends. The two chambers into which it is divided contain hydrogen and oxygen. The two diaphragms are set in vibrations of same frequency. What is the minimum frequency of these diaphragms for which the middle diaphragm will be motionless? Velocity of sound in hydrogen is `1100 m//s` and that in oxygen is `300 m//s`.

Explanation

As diaphragm C is a node, A and B will be antinodes (as in organ pipe either both ends are antinode or one end node and the other antinode), i.e., each part will behave as a closed end organ pipe so that
`f_(H)=(v_(H))/(4L_(H))=(1100)/(4xx0.5)=550 Hz and`
`f_(0)=(v_(0))/(4L_(0))=(330)/(4xx0.5)=150Hz`
As the two fundamental frequencies are different, the system will vibrate with a common frequency `f_(C)` such that `f_(C)=n_(H)f_(H)=n_(0)f_(0), " ". (n_(H))/(n_(0))=(f_(0))/(f_(H))=(150)/(550)=(3)/(11)`
Then the third harmonic of hydrogen and 11th harmonic of oxygen or 9 th harmonic of hydrogen and 33rd harmonic of oxygen will have same frequency. So the minimum common frequency.
`f=3xx550 or 11xx150 Hz`
(as 6th harmonic of H and 22nd of O will not exist.)