Question

A wire of density $9\times {10}^{-3}{\text{kg cm}}^{-3}$ is stretched between two clamps $1\text{m}$ apart. The resulting strain in the wire is $4.9\times {10}^{-4}$. The lowest frequency of the transverse vibrations in the wire (in $\text{Hz}$) is

(Young's modulus of wire $Y=9\times {10}^{10}{\text{Nm}}^{-2})$

Answer 1

Fundamental frequency of vibration of the string,

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

We can rewrite the above equation as,

$f=\frac{1}{2l}\sqrt{\frac{T}{\rho A}}$

$f=\frac{1}{2l}\sqrt{\frac{Y\Delta l}{\rho l}}$

[Using $Y=\frac{T/A}{\Delta l/l}$]
From the data given,

$f=\frac{1}{2\times 1}\sqrt{\frac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{9\times {10}^{-3}\times {10}^6}}=35.00\text{Hz}$