A steel wire of length 1 m, mass 0.1 kg and uniform cross-sectional area `10^(-6)m^(2)` is rigidly fixed at both ends. The temperature of wire is lowered by 20°C. If transverse waves are set up by plucking the string in the middle, calculate the frequency (In S.I. units) of the fundamental mode of vibration. Young’s modulus of steel `=2xx10^(11)N//m^(2)`, coefficient of linear expansion of steel`=1.21xx10^(-6)(degC)^(-1)`.

Explanation

The mechanical strain `=(trianglel)/(l)=alphatriangleT=1.21xx10^(-5)xx20=2.42xx10^(-5)`
The tension in wire `=T=Y(trianglel)/(l)A=2xx10^(11)xx2.42xx10^(-5)xx10^(-6)=48.4N`
`therefore` speed of wave in wire
`V=sqrt((T)/(mu))=sqrt((48.4)/(0.1))=22m//s`
Since the wire is plucked at `(l)/(4)` from one end the wire shall oscillate in `1^(st)` overtone (for minimum number of loops)
`lamda=l=1m`
Now `V=flamda` or `f=(V)/(lamda)=22Hz`