A steel wire of length `50sqrt(3) cm` is connected to an aluminium wire of length `60 cm` and stretched between two fixed supports. The tension produced is `104 N`, if the cross section area of each wire is `1mm^(2)`. If a transverse wave is set up in the wire, find the lowest frequency for which standing waves with node at the joint are produced . (density of aluminimum `= 2.6 gm//cm^(3)` and density of steel `= 7.8 gm//cm^(3)`).

Answer

`1000//3Hz`

Explanation

Steel , Aluminium
`T = 104`
`sigma_(1) = 7.8 gm//cm^(3) , sigma_(2) = 2.6 gm//cm^(3)`
`A_(1) = A_(2) = 1 mm^(2)`
`V_(1) = sqrt((T)/(sigma_(1)A_(T))) = sqrt((104)/((7.8 xx 10^(-3))/(10^(-6)) xx 1 xx 10^(-6)))`
`V_(1) = ((200)/(sqrt(3))) m//sec`.
`v_(2)sqrt((T)/(sigma_(2)A_(2))) = sqrt((104)/(2.6 xx 10^(-3)))`
`= 2 xx 10^(2) m//sec`.
for lowest freq.
`f_(1) = f_(2)`
`(n_(1)V_(1))/(2L_(1)) = (n_(2)V_(2))/(2L_(2))`
`(n_(1) xx 200)/(2 xx sqrt(3) xx 50sqrt(3)) = (n_(2) xx 200)/(2 xx 60 xx 1)`
`rArr [(n_(1))/(n_(2)) = ((5)/(2))]`
So lowest freq.
`f_(2) = (n_(2)V)/(2L) = (2 xx 200)/(2 xx 0.6) = (1000)/(3)Hz`