A wire of density `9 xx 10^(3) kg//m^(3)` is stretched between two clamps 1 m apart and is subjected to an extension of `4.9 xx 10^(-4) m`. The lowest frequency of transverse vibration in the wire is `(Y = 9 xx 10^(10) N//m^(2))`

Explanation

Let A be the area of cross-section ofwire and T be the tension applied
The lowest frequency oftransverse vibration is given by `n = (1)/(2l) sqrt((T/mu))`
Where m = mass per unit length of wire
= Volume of unit length xx density
= A xx 1 xx p
`n = (1)/(2l) sqrt((T/(Ap)))`
`y = (stress)/("strain") = (TIA)/(Deltalll)`
`T = YA ((Delta l)/(l))`
From equations-(6.183) and (6.184)
`n = (1)/(2l) sqrt(((YA Deltal)/(l A p))) = (1)/(2l) sqrt(((YDeltal)/(lp)))` Substituting the given values
`n = (1)(2xx1) sqrt((((9xx10^(10)) xx (4.9 xx 10^(-4)))/(1 xx (9xx10^(3)))))`
= 35 Hz