A piano wire with a diameter of 0.90 ~mm is replaced by another wire of diameter 0.93 ~mm of the same material. If tension of wire is kept the same, then the percentage change in frequency of fundamental tone is

Step-by-step solution

Given, diameter, d₁=0.9 ~mm =0.9 10⁻³ ~m aligned & =9 10⁻⁴ ~m d₂=0.93 ~mm & =9.3 10⁻⁴ ~m aligned Frequency of fundamental tone is given as f= 1 2 l T m where, m= mass per unit length. = M l = V l = A l l =A m= d^2 4 From Eqs. (i) and (ii), we get aligned & f= 1 2 l T d^2 / 4 = 1 2 l 4 T d^2 = 1 l T d^2 & f= 1 l d T f 1 d & f₁ f₂ = d₂ d₁ = 9.3 10⁻⁴ 9 10⁻⁴ = 9.3 9 f₂ f₁ = 9 9.3 aligned Percentage change in frequency aligned & = f₂-f₁ f₁ 100= ( f₂ f₁ -1 ) 100 & = ( 9 9.3 -1 ) 100=-3.2 % aligned