A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is `1.0 mm^2`, find its Young modulus.

Answer

1.985xx10^11N/m^2

Explanation

`L=40cm=.4m`
mass=3.2g=3.2x10^-3kg`
`:. Mass per unit length
`m=(3.2xx10^-3)/0.4=8xx10^-3(kg)/m`
Change in length
`/_\\L=40.05=80xx10^-3(kg)/m`
Strain` =(/_\\L)/L=((0.05xx10^-2))/0.4`
`=0.125xx10^-2`
`f=220Hz`
`f=1/(2L)sqrt(T/m)`
`=1/(2xx(0.4005))`
`sqrt((T/(8xx10^-3)))`
`rarr 220xx220=[1/((0.801)^2)]xxTxx{10^3/8)`
`rarr Txx1000=220xx220xx0.641xx0.8`
`rarr T=248.19N`
Strain =`248.19/1(1mm^2)`
Strain =`248.19/(1mm^2)`
`=248.19/10^-6=248.19xx10^6`
`Y`=stress/strain
`=((248.19xx10^6))/((0.125xx10^-2))
=1985.52xx10^8
=1.985xx10^11N/m^2`