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An aluminium wire of cross-sectionbal area 1 xx 10^(-6)m^(2) is joined to a steel wire of same cross-sectional ara. This compound wire is stretched on a sonometer, pulled by a weight of 10 kg. The total length of the compound wire between the bridges is 1.5 m, of which the aluminium is 0.6 m and the rest is steel wire. Transverse vibrationas are set up in the wire by using an ecternal force of variable frequency. Find the lowest frequency of excitation for which standing waves are formed such that the joint in the wire is a node. What is the total number of nodes observed at this frequency, excluding the two at the ends of the wire? The density of aluminium is 2.6 xx 10^(3) kg//m^(3) and that of steel is 1.04 xx 10^(4) kg//m^(3). |
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Answer» `sigma_(1) = 2.6 xx 10^(3)kg//m^(2) , sigma_(2) = 1.04 xx 10^(4) kg//m^(3)` `v_(1) = sqrt(100/(1xx10^(6)xx1.04xx10^(3)))=(10^(3))/(sqrt(26))m//sec` `V_(2)=sqrt(100/(1xx10^(-6)xx1.04xx10^(4)))=(10^(3))/(2sqrt(26))` `f_(1)=f_(2)` `(n_(1)xx10^(3))/(2 xx 0.6sqrt(26)) = (n_(2) xx 10^(3))/(2 xx 2 xx 0.9sqrt(26)) [(n_(1))/(n_(2)) = (1)/(3)]` Lowest FREQUENCY `= (n_(1)v_(1))/(2L_(1)) = (1 xx 10^(3))/(sqrt(26) xx 2 xx 0.6)` `= (10^(4))/(12sqrt(26)) = 162 "vibration"//"sec"`. `(n_(1))/(n_(2)) = ((1)/(3))` One side `1` loop and other side `3` loop  So EXCLUDING the two nodes at ends total nodes is `3` 
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