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A steel wire of length 1 m, mass 0.1 kg ...

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)`.

Text Solution

Verified by Experts

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
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