Question

AB is a cylinder of length 1m fitted with a thin flexible diaphragm C at middle and two other thin flexible diaphragms A and B at the ends as shown. The portions AC and BC contain hydrogen and oxygen gases respectively. The diaphragms A and B are set into vibrations of the same frequency. The minimum frequency of these vibrations for which diaphragm C is a node, is (Under the conditions of the experiment the velocity of sound in hydrogen is 1100 m/s and oxygen 300 m/s)

• A. 1100 Hz
• B. 3300 Hz
• C. 1650 Hz
• D. 1500 Hz

Answer 1

The correct option is C 1650 Hz

As diaphragm C is a node, and A and B are antinodes, each part behaves as closed organ pipe.

Thus fundamental frequency of the sections are-

$n_H=\frac{{\nu }_H}{4l_H}=\frac{1100}{4\times 0.5}=550Hz$

$n_O=\frac{{\nu }_O}{4l_O}=\frac{330}{4\times 0.5}=150Hz$.

Since their fundamental frequencies are unequal, we need to find those overtones when they are resonating.

Thus $n=p{n}_H=q{n}_O$

$⟹\frac{p}{q}=\frac{3}{11}$

Smallest integer value of p is 4.

Thus the minimum frequency of vibration of diaphragm C is

$n=3n_H=1650Hz$