`AB` is a ccylinder of length `1m` fitted with a thin flexible diaphragn `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`).

Answer

A

Explanation

It is given that `C` acts as a node. This implies that at `A` abd `B` antitodes are formed. Again it is given that the frequencies are same.

rArr `(v_(1))/(4l) xx p_(1) = (v_(2))/(4l) xx p_(2)` or `(P_(1))/(P_(2)) = (v_(1))/(v_(2)) = (3)/(11)`
or, `11p_(1) = 3p_(2)`
This means that the third harmonic in `AC` is equal to `11th` harmonic in `CB`.
Now, the fundamental frequency in `AC`
`= (v_(1)/(4l) = (1100)/(4 xx 0.5) = 550 Hz`
:. Frequency in `AC = 3 xx 550 = 1650 Hz` and frequency in `CB = 11 xx 150 = 1650Hz`