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 wm be antinodes (as is an organ pipe, either both ends are antlnode or one end node and the other antinode), i.e., each part will behave as a closed end organ pipe that
`f_H = v_H/(4L_H) =1100/(4 xx 0.5) = 550 Hz` and
`f_0 =v_0/(4L_0) = 300/(4 xx 0.5) =150 Hz`
As the two fundamental frequencies are different,the system will vibrate with a common frequency `f_C` such that `f_C =n_Hf_H`
`n_0 f_0 rArr n_H/n_0 = f_0/f_H = 150/550 =3/11`
Then the third harmonic of hydrogen and `11^(th)` harmonic of oxygen or `9^(th)` harmonic or hydrogen and 33rd harmonic of oxygen will have same frequency. So the minimum common frequency.
`f = 3 xx 550 or 11 xx 150 Hz = 1650 Hz`
(as `6^(th)` harmonic of H and `22^(nd)` of O will not exist.)