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AB is a cylinder of length 1 m fitled with an inflexible diaphragm C at middle and two other thin flexible diapharogms A and B at the ends. The portions AC and BC maintain hydrogen and oxygen gases respectively. The diaphragms A and B are sel into vibrations of the same frequency. What is the minimum frequency of these vibralions for which diaphragm C is a 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 wm be antinodes (as is an organ pipe, either both ends are antlnode orone end node and the other antinode), i.e., each PART will behave as a closed end organ pipe that `f_H = v_H/(4L_H) =1100/(4 xx 0.5) = 550 Hz` and `f_0 =v_0/(4L_0) = 300/(4 xx 0.5) =150 Hz` As the two fundamental frequencies are different,the system will vibrate with a common FREQUENCY `f_C` such that `f_C =n_Hf_H` `n_0 f_0 rArr n_H/n_0 = f_0/f_H = 150/550 =3/11` Then the third harmonic of hydrogen and `11^(th)` harmonic of oxygen or `9^(th)` harmonic or hydrogen and 33rd harmonic of oxygen will have same frequency. So the minimum common frequency. `f = 3 xx 550 or 11 xx 150 Hz = 1650 Hz` (as `6^(th)` harmonic of H and `22^(nd)` of O will not EXIST.) |
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