An aluminium wire and a steel wire of same cross-sectional of `10^(-2)cm^(2)` were connected together. A 10 kg block loaded on the compound wire as shown in figure. Waves are set up in the compound wire. The minimum frequency of excitation for which standing waves are observed with point A as a node is `[L_(1)=60cm,L_(2)=86.6cm]` density of aluminium `=2.6g//cm^(3)` and of steel=`7.8g//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`