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# . AB is a cylinder of length 1 m 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 diaphragm C is a node? (Under the conditions of experiment

7 years ago
Hello Student,
It is given that C as a node. This implies that at A and B antinodes are formed. Again it is given that the frequencies are same.
$\Rightarrow$v1/4ℓ x p1 = v2/4ℓ x p2 or p1/p2 = v1/v2 3/11
Or, 11p1 = 3p2
This means that the third harmonic in AC is equal to 11th harmonic in CB.
Now, the fundamental frequency in AC
= v1 4ℓ = 1100/4 x 0.5 = 550Hz
And the fundamental frequency in CB
= v2/4ℓ = 300/4 x 0.5 = 550 Hz
∴ Frequency in AC = 3 x 550 = 1650 Hz
And frequency in CB = 11 x 150 = 1650Hz.

Thanks