A steel wire of length 1.25 ~m is stretched between two rigid supports. The tension in the wire produces an elastic strain of 0.14 % . The fundamental frequency of the wire is (Density and Young's modulus of steel are 7.7 10^3 kgm ⁻³ and 2.2 10¹¹ Nm ⁻² respectively)

Step-by-step solution

Given, length of wire =1.25 ~m strain produced =0.14 % density of wire, d=7.7 10^3 kgm ⁻³ young's modulus, Y=2.2 10¹¹ Nm ⁻² Fundamental frequency of vibration over a string is given by f= 1 2 L T Where, T= Tension in string, L= Length of string and = mass per unit length of string. So, we can express frequency as f= 1 2 L T / A M / l A From young's modulus of elasticity we have, aligned & & Y= T / A l / l & T A & =2.2 10¹¹ 0.14 100 & T A & =0.308 10^9 aligned Substituting for L, T A and d in eq. (i) we get, Fundame