An aluminium wire of cross-sectional area `(10^-6)m^2` is joined to a steel wire of the same cross-sectional area. This compound wire is stretched on a sonometer pulled by a weight of 10kg. The total length of the compound wire between the bridges is 1.5m of which the aluminium wire is `0.6 m` and the rest is steel wire. Transverse vibrations are setup in the wire by using an external source of variable frequency. Find the lowest frequency of excitation for which the standing waves are formed such that the joint in the wire is a node. What is the total number of nodes at this frequency? 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^2 (g = 10m//s^2)`

Explanation

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)`