An aluminimum wire of cross - sectional area ` 1 xx 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 `10 kg`. The total length of the compound wire between the bridges is `1.5 m` of which the aluminium wire is `0.6 m` and the rest is steel wire . Transverse vibrations are set up in the wire by using an external source 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)`

Explanation

Let `l_(1) and l_(2)` be the lengths of aluminum and steel wires , respectively .
Given ` l _(1) = 0.6 m , l_(1) + l_(2) = 1.5 m`
`:. L_(2) = 1.5 - 0.6 = 0.9 m`
If aluminium wire vibrates in `p_(1)` loops and steel wire in `p_(2)` loops , then frequency `n` is given by
`n = n_(1) = (P_(1))/( 2l_(1)) sqrt (((T)/( m_(1)))) = (P_(1))/( 2l_(1)) sqrt (((T)/(A rho_(1))))`
Also , `n = n_(2) = (P_(2))/( 2l_(2)) sqrt (((T)/( m_(2)))) = (P_(2))/( 2l_(2)) sqrt (((T)/(A rho_(2))))`
`:. (P_(1))/(P_(2)) = (l_(1))/( l_(2)) sqrt(((rho_(1))/( rho_(2)))) = ( 0.6)/(0.9) sqrt(((2.6 xx 10^(3))/(1.04 xx 10^(4))))`
`:. P_(1) : P_(2) = 1 : 3`
The minimum number of loops in aluminimum wire ` = 1` and minimum number of loops in steel wire `= 3`.
The stationary vibratuions in composite wire as shown in Fig. 7.69. Obviously total number of nodes `= 5`.
The number of nodes excluding the two at the ends ` = 5 - 2 = 3` . Thus , the lowest frequency
` n = (1)/( 2 l_(1)) sqrt(((T)/( m_(1)))) = (1)/( 2 xx 0.6) sqrt (((10 xx 9.8)/( 1 xx 10^(-6) xx 2.6 xx 10^(3)))) = 162 Hz`