question_answer

AB is a cylinder of length 1.0 m filled with a thin flexible diaphragm C (see figure) at the middle and two other thin flexible diaphrams A and B at the ends. The portions AC and BC contain hydrogen and oxygen gases respectively. The diaphrams A and B are set into vibrations of same frequency. What is the minimum frequency of these vibrations for which the diaphrams C is a node? Under the conditions of the experiment, the velocity of sound in hydrogen is \[1100 m/s\]and in oxygen is \[300 m/s.\]

A) \[1650Hz\]        

B) \[1560Hz\]

C) \[650Hz\]                      

D) \[560Hz\]

Correct Answer: A

Solution :

[a]

As ends \[A\] and \[B\] are set in vibrations, so displacement antinodes are formed at these ends. The fundamental

Frequency of each pipe is corresponding to one node and one antinode. If \[{{f}_{1}}\operatorname{and}\,{{f}_{2}}\] be the fundamental frequencies of gases in \[AC\] and \[BC\] respectively, then \[{{f}_{1}}=\frac{{{v}_{1}}}{4L}=\frac{1100}{4\times 0.5}=550\operatorname{Hz}\] and \[{{f}_{2}}=\frac{{{v}_{2}}}{4L}=\frac{300}{4\times 0.5}=150\operatorname{Hz}\]

As the two frequencies are different, so the two columns are not vibrating in the fundamental mode. The close column of gas vibrates only in odd harmonics with frequencies \[1:3:5:7:.......\]

Thus we can write\[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{550}{150}=\frac{11}{3},\frac{22}{6},\frac{33}{9}\] or \[3{{f}_{1}}=11{{f}_{2}}\]

The common minimum frequency \[=3{{f}_{1}}=3\times 550=1650\operatorname{Hz}\]

Also \[=11{{f}_{2}}=11\times 150=1650\operatorname{Hz}\]