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

Answer

A

Explanation

Fundamental frequency of vibration of sonometer wire is `f_(0)=(1)/(2L)sqrt((T)/(mu))`
where L is the length of the wire, T is the tension in the wire and is the mass per length of the wire
As `mu=pA`
Where p is the density of the material of the wire and A is the area of cross-section of the wire.
`therefore` `f_(0)=(1)/(2L)sqrt((T)/(pA))` But we have `Y=(Stress)/(Stra i n)=(TL)/(ADeltaL)`
The tension in the wire is due to elasticity of wire
`therefore` `T=YA[(DeltaL)/(L)],` Hence, `f_(0)=(1)/(2L)sqrt((Y Delta L)/(pL))`
`Y=2.2xx10^(11)N//m^(2),p=7.7xx10^(3) kg//m^(3),(DeltaL)/(L)=0.01,L=1.5m`
Substituting the given values, we get : `f_(0)=(1)/(2xx1.5)sqrt((2.2xx10^(11)xx0.01)/(7.7xx10^(3)))=(10^(3))/(3)sqrt((2)/(7))Hz=178.2 Hz`