A sonometer wire of length `1.5m` 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 xx 10^(3) kg//m^(3)` and `2.2 xx 10^(11) N//m^(2)` respectively ?

Options

A. 178.2 Hz
B. 200.5 Hz
C. 770 Hz
D. 188.5 Hz

Answer

A

Explanation

T is the tension in the wire and `mu` is the mass per unit length.
When it is opend at both ends fundamental frequency is
`f = (upsilon)/(2l) = (1)/(2l) sqrt((T)/(m)) = (1)/(2l) sqrt((T)/(Ad))" "...(i)`
Yound modulus of string is
`Y = (Tl)/(A Delta l) rArr (T)/(A) = (Y Delta l)/(l)" "...(ii)`
From equation (i) and (ii)
`rArr" "f = (1)/(2l) sqrt((Y Delta l)/(ld))`
Given, l = 1.5 m, `(Delta l)/(l) = 0.01, d = 7.7 xx 10^(3) kg//m^(3)`
`Y = 2.2 xx 10^(11) N//m^(2)`
`f = sqrt((2)/(7)) xx (10^(3))/(3)` Hz
`f ~~ 178.2` Hz