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 is (Young’s modulus of wire $Y=9\times {10}^{10}{\text{Nm}}^{–2})$, (to the nearest integer),

Answer 1

Given, Denisty of wire, $\sigma =9\times {10}^{-3}{\text{kg cm}}^{-3}$

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

Strain$=4.9\times {10}^{-4}$

$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{T/A}{\text{Strain}}$
$\therefore \frac{T}{A}=Y\times \text{Strain}=9\times {10}^{10}\times 4.9\times {10}^{-4}$

Also, mass of wire, $m=Al\sigma$

Mass per unit length, $\mu =\frac{m}{J}=A\sigma$
Fundamental frequency in the string
$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}=\frac{1}{2l}\sqrt{\frac{T}{\sigma A}}=\frac{1}{1\times 2}\sqrt{\frac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{9\times {10}^3}}$
$\frac{1}{2}\sqrt{49\times {10}^{9-4-3}}=\frac{1}{2}\times 70=35\text{Hz}$.