The speed of a transverse wave going on a wire having a length 50 cm and maas 5.0 g is `80ms^(-1)` The area of cross section of the wire is `1.0mm.^2` and its Young modulus is `16 xx 10^(11) N m^(-2).` Find the extension of the wire over its natural lenth.

Answer

0.02

Explanation

The linear mass density is `mu=(5xx10^(-3)kg)/(50xx10^(-2)m)=1.0xx10^(-2)(kg)/(m)`
The wave speed is `V=sqrt((F//mu)`
Thus, the tension is
`F=muv^(2)=(1.0xx10^(-2)(kg)/(m))xx6400(m^(2))/(s^(2))=64N`.
Young's modulus is given by `Y=(F//A)/(DeltaL//L)`.
Therefore extension is, `DeltaL=(FL)/(AY)=((64N)(0.50m))/((1.0xx10^(-6)m^(2))xx(16xx10^(11)N//m^(2)))=0.02m`m