A `40 cm` long wire having a mass `3.2 g` and area of cross - section ` 1 mm^(2)` is stretched between the support `40 .05 cm` apart. In its fundamental mode , it vibrates with a frequency `1000//64 Hz`. Find the young's modulus of the wire.

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)`