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

A and B are set in oscillation, hence antinodes are situated at these points. The portions of the cylinders i.e., AC and CB behaves as a close cylinder at the end C. Thus there is immediate node at C. The fundamental frequency of each pipe corresponds to just one node and one anti node. Let `n_(1)` and `n_(2)` be the fundamental frequencies of gases in AC and BC respectively, then
`n_(1) = (v_(1))/(4l) = (1100)/(4 xx 0.5) ` = 550 Hz
`AC = l = 0.5`
`n_(2) = (v_(2))/(4l) = (300)/(4 xx 0.5)` = 150 Hz
Here `v_(1)` and `v_(2)` are the velocities of sound in hydrogen and oxygen resectively.
As the two frequencies are different and hence the two gas columns are not vibrating in the fundamental mode. In case ofa closed pipe only odd harmonics with frequencies 3,5,7,9,.. etc. times the fundamental frequency are observed. Now the problem is to find out that which harmonics of `n_(1)` and `n_(2)` have the same frequency. We notice that
`(n_(1))/(n_(2)) = (500)/(150) = (11)/(3)`
`3n_(1)` = `11n_(2)`
This shows that third harmonic of `n_(1)` and eleventh harmonic of `n_(2)` have equal frequencies. Similarly, 6th harmonic of `n_(1)` and 22th harmonic of `n_(2)` have equal frequencies and so on.
Thus common minimum frequency is
= `3n_(1) = 3 xx 500 = 1650 Hz`
= `11n_(2)` = 11 xx 150 = 1650Hz`.