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. `188.5` Hz
B. `178.2` Hz
C. `200. 5` Hz
D. 770 Hz

Answer

b

Explanation

Fundamental frequency of sonometer wire
`f=v/(2l)=1/(2l)sqrt(T/mu)=1/(2l)sqrt(T/(Ad))`
Here, `mu=` mass per unit length of wire.
Also, Young's modulus of elasticity `Y=(Tl)/(ADeltal)`
`rArr T/A=(YDeltal)/l`
`rArr f=1/(2l)sqrt((YDeltal)/(ld))`
`l= 1.5 m, (Deltal)/l=0.01, d-7.7xx10^(3)kg M^(-3)`
`y=2.2xx10^(-2)`
After substituting the values, we get, `fapprox 178.2` Hz