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An aluminium wire of cross-sectional area 10^(-6)m^2 is joined to a copper wire of the same cross-section. This compound wire is stretched on a sonometer, 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 up in the wire in the lowest frequencyof excitation for which standing waves are formed such that the joint in the wire is a node. What is the total number of nodes qbserved at this frequncy excluding the twoat the endsofthe wire? Thedensity of aluminium is 2.6 xx 10^4 kg//m^3. |
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Answer» Solution :As the total LENGTH of the wire is 1.5 m and out of which `L_A =0.6m`, so the length of copper wire `L_c = 1.5-0.6 =0.9 m`.The tension in the whole wire is same `(= Mg =10g N)` and as FUNDAMENTAL frequency of vibrationof string is given by `f =1/(2L) sqrt(T/m) =1/(2L) sqrt(T/(rho A))` [ as ` m =rhoA`] So `f_A =1/(2L_A) sqrt(T/(rho_A A) and f_c = 1/(2L_c) sqrt(T/(rho_cA)`......(1) Now as in case of composite wire, the whole wire will vibrate with fundamental frequency `f = n_A f_A =n_c f_c`......(2) Substituting the values of `f_A and f_c` from Eqn. (1) in (2) `n_A/(2 xx 0.6) sqrt(T/(A xx 2.6xx10^3)) = n_c/(2 xx 0.9) sqrt(T/(A xx 1.0401 xx 10^4))` i.e., `n_A/n_c =2/3 sqrt(2.6/10.4) = 2/3 xx 1/2 = 1/3` So that for fundamental frequency ofcomposite string,`n_A = 1 and n_c =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` Thsi 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 =2/(2 xx0.6) sqrt((10xx9.8)/(10^(-6) xx 2.6 xx 10^3)) ~~ 161 .8 Hz (= 3f_c)` |
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