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Question

A wire of density 9×103 kg 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 (in Hz) is

(Young's modulus of wire Y=9×1010 Nm2)

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Solution

Fundamental frequency of vibration of the string,

f=12lTμ

We can rewrite the above equation as,

f=12lTρA

f=12lYΔlρl

[Using Y=T/AΔl/l]
From the data given,

f=12×19×1010×4.9×1049×103×106=35.00 Hz

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