A steel wire of length `1m`, mass `0.1kg` and uniform cross-sectional area `10^(-6)m^(2)` is rigidly fixed at both ends. The temperature of the wire is lowered by `20^(@)C`. If transverse waves are set up by plucking the string in the middle.Calculate the frequency of the fundamental mode of vibration.
Given for steel `Y = 2 xx 10^(11)N//m^(2)`
`alpha = 1.21 xx 10^(-5) per ^(@)C`

Explanation

Coefficient of linear expansion is given by
`alpha=(Deltal)/(lxxDeltatheta)` where `Deltatheta` = change in temperature
Hence internal strain= `(alpha.Deltatheta)`
`therefore` Stress = Y x strain = `Y alpha Delta theta`
Tension `T=pi r^2 xx "stress" = pir^2 Y alpha Deltatheta`
`therefore v=sqrt(T/m)`
`therefore n=1/(2l) sqrt(T/m) =1/(2l) sqrt((pir^2 YalphaDeltatheta)/m)=1/(2xx1)xxsqrt((10^(-6)xx2xx10^11 xx1.21xx10^(-6)xx20)/0.1)` =11Hz