Question

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)

Answer 1

It is given that c as a node. this implies that A and B antinodes are formed. Again it is given that the frequencies are same.

p1/p2 = v1/v2 3/11

Or, 11p1 = 3p2

this means that the third harmonic in AC is equal to 11th harmone in CB.

Now the fundamental frequency in AC= 1100/4 * 0.5= 550Hz

And the fundamental frequency in CB= 300/4 * 0.5= 150Hz

Frequency in AC= 3*550= 1650Hz

Frequency in CB= 11*150 = 1650Hz