Question

AB is a cylinder of length 1 m fitted with a thin flexible diaphgram C at middle and
two other thin flexible diaphgrams A and B at the ends. The portions AC and BC maintain hydrogen and oxygen gases respectively. The diaphgrams A and B are set into vibrations of the same frequency. What is the minimum frequency of these.vibrations for which diaphgram C is a node? Under the conditions of the experiment velocity of sound in hydrogen is 1100 m/s and oxygen 300 m/s.​

Answer 1

AnSwer :-

As diaphragm C is node, A and B will be antinodes (as in organ pipe either both ends are antinodes or one end node and the other antinode) ,i.e., each part will behave as a closed end organ pipe so that

$\sf f_H = \dfrac{v_H}{ 4L _H }= \dfrac{1100}{4 \times 0.5} = 550Hz$

and,

$\sf f_0 = \dfrac{v_0}{4L_0} = \dfrac{330}{4 \times 0.5} = 150Hz$

As the two fundamental frequencies are different, the system will vibrate with a common frequency f$\sf _c$ such that

$\sf f_c = n_Hf_H = n_0f_0$

$\sf \dfrac{n_H}{n_0}=\dfrac{f_0}{f_H} = \sf \dfrac{150}{550}=\dfrac{3}{1}$

Then the third harmonic of Hydrogen and 11th harmonic of oxygen or 9th harmonic of Hydrogen and 33rd harmonic of oxygen will have same frequency. so the minimum Common frequency

f = 3 × 550 or 11 × 150 Hz

[as 6th harmonic of H and 22nd of O will no exist ]

Thus the minimum frequency of vibration of diaphragm C is

n = 3n$\sf _H$ = 1650Hz

$\:$

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