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Question

A wire of density 9×103kg cm3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9×104. The lowest frequency of the transverse vibrations in the wire is (Young’s modulus of wire Y=9×1010 Nm2), (to the nearest integer),

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Solution

Given, Denisty of wire, σ=9×103 kg cm3

Young's modulus of wire,Y=9×1010 Nm2

Strain=4.9×104

Y=StressStrain=T/AStrain
TA=Y×Strain=9×1010×4.9×104

Also, mass of wire, m=Alσ

Mass per unit length, μ=mJ=Aσ
Fundamental frequency in the string
f=12lTμ=12lTσA=11×29×1010×4.9×1049×103
1249×10943=12×70=35 Hz.

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