A sonometer wire of length 1.5m is made of steel. The tension in it produce an elastic strain of 1%. What is the fundamental frequency of transverse vibrations in Hz if density and young’s modulus of steel are  $7.7\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$ and $15.4\times {10}^{11}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$ respectively? $\left(\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236\right)$

Given length of wire  $l=1.5\text{\hspace{0.17em}m}$
Young’s modulus  $Y=\frac{\text{stress}}{\text{strain}}=15.4\times {10}^{11}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$
Stress=Tension/Cross section area
 = $\frac{T}{A}$
Strain=1%=0.01
$\Rightarrow 15.4\times {10}^{11}=\frac{T/A}{0.01}$

$\Rightarrow 15.4\times {10}^{11}=\frac{T}{0.01A}$

$\Rightarrow T=15.4\times {10}^9\times A$

Volume density  $d=7.7\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$
Linear density  $\mu =\text{Volume\hspace{0.17em}density×Area}$
$=d\times A$      
$=7.7\times {10}^3\times A$     
Velocity of transverse wave on the sonometer wire $V=\sqrt{\frac{T}{\mu }}$
$\Rightarrow V=\sqrt{\frac{15.4\times {10}^9}{7.7\times {10}^3}}$ $=\sqrt{2}\times {10}^3\text{m}/\text{s}$ 
Fundamental note has a wave length  $\lambda =2L=3\text{\hspace{0.17em}m}$
So, frequency= $\frac{\text{V}}{\lambda }=\frac{\sqrt{2}\times {10}^3}{3}=471.33\text{Hz}$.