Discussion Forum

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 x10³ kg/m³ and 2.2 x 10¹¹ N/m²  respectively?

Answer:

Explanation:

Fundamental frequency of sonometer wire

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

                               $\frac{1}{2l}\sqrt{\frac{T}{Ad}}$

 Here,  $\mu$ = mass per unit length of wire,

Also , Young's modulus of elesticity

             $Y=\frac{Tl}{A\triangle l}$

$\Rightarrow\frac{T}{A}=\frac{Y\triangle l}{l}\Rightarrow f=\frac{1}{2l}\sqrt{\frac{Y \triangle l}{ld}}$

l=1.5 m, d=7.7 x 10³ kg/m³

  Y= 2.2 X 10¹¹  N/m²

 After substituting the values we get,

 f≈ 178.2 Hz