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

`10^(9) N//m^(2)`

Explanation

For fundamental frequency
`mu = ( 3.2 g)/(40 cm) = ( 3.2 xx 10^(-3))/( 40 xx 10^(-2))`
` = (32)/(4000) kg//m`
`l = lambda//2`
`lambda = 2l`
`f = (v)/( lambda) = (1)/( 2 l) sqrt((T)/( mu))`
`(1000)/(64) = (1)/( 2 xx 40 xx 10^(-2)) sqrt((T)/(32 //4000))`
`[ (1000)/(64) xx 2 xx 40 xx 10^(-2)]^(2) (32)/(4000) T`
`rArr T = (10)/( 8) N`
`Y = ( stress)/("strain") = ((10//8)/(10^(-6)))/((0.05 xx 10^(-2))/(40 xx 10^(-2))) = 10^(9) N//m^(2)`