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AB is a cylinder of length I m fitted with a thin flexible diaphragm Cat middle and two other thin flexible diapharigms A and B at the ends. The portions AC and BC maintain hydrogen and oxygen gases respectively. The diaphragms A and B are set into vibrations of the same frequency. What is the minimum frequency of these vibrations for which diaphragm Cisa node? Under the conditions of the experiment velocity of sound in hydrogen is 1100 m/s and oxygen 300 m/s |
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Answer» Solution :As diaphragm C is a node, A and B will be antinodes (as in organ PIPE EITHER both ends are ANTINODE or one end node and the other antinode), i.e., each part will behave as a closed end organ pipe so that `f_(H)=(v_(H))/(4L_(H))=(1100)/(4xx0.5)=550 Hz and` `f_(0)=(v_(0))/(4L_(0))=(330)/(4xx0.5)=150Hz` As the two fundamental frequencies are different, the system will vibrate with a common FREQUENCY `f_(C)` such that `f_(C)=n_(H)f_(H)=n_(0)f_(0), "". (n_(H))/(n_(0))=(f_(0))/(f_(H))=(150)/(550)=(3)/(11)` Then the third harmonic of hydrogen and 11th harmonic of oxygen or 9 th harmonic of hydrogen and 33rd harmonic of oxygen will have same frequency. So the minimum common frequency. `f=3xx550 or 11xx150 Hz` (as 6th harmonic of H and 22nd of O will not exist.) |
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