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An aluminium wire of cross-sectional area 10"mis joined to a copper wire of the same crosssection. This compound wire is stretched from a fixed end, pulled by a load of 10 kg. The total length of the compound wire between two bridges is 1.5 m of which the aluminium wire is 0.6 m and the rest is the copper wire. Transverse vibrations are set in the wire by rising an external force of variable frequeney. Find the lowes! frequency of excitation for which standing waves are formed, such that the joint in the wire is anode. What is the total number of modes observed at this frequency excluding the two at the ends of the wire? The density of aluminium is 2.6xx10^(4) kg//m^(3). |
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Answer» Solution :Now as in case of composite wire, the WHOLE wire will vibrate with fundamental frequency. `f=n_(A)f_(A)=n_(c)f_(c)` `(n_(A))/(2xx0.6) sqrt((T)/(Axx2.6xx10^(-3)))=(n_(c))/(2xx0.9) sqrt((T)/(Axx1.0401xx10^(4)))` `i.e (n_(A))/(n_(e))=(2)/(3) sqrt((2.6)/(10.4))=(2)/(3)xx(1)/(2)=(1)/(3)` So that for fundamental frequency of composite string, N = 1 and n = 3, i.e., ALUMINIUM string will vibrate in first harmonic and copper wire at second, overtone as shown in figure.  `:.f=f_(A)=3f_(c)` This in turn implies that total NUMBER of nodes in the string will be 5 and so number of nodes excluding the nodes at the ENDS =5-2=3 and `f=f_(A)=(1)/(2xx0.6) sqrt((0xx9.8)/(10^(-6)xx2.6xx10^(3)))~~161.8 Hz (=3f_(c))` |
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