When `1 m` long metallic wire is stressed, an extension of `0.02 m` is produced. An organ pipe `0.5 m` long and open at both ends, when sounded with this stressed metallic wire, produced `8` beats in its fundamental mode. By decreasing the stress in the wire, the number of beats are found to decrease. Find the Young's modulus of the wire. The density of metallic wire is `10^(4) kg//m^(3)` and velocity of sound in air is `292 m//s`.

Explanation

`L = 1 m, Delta L = 0.02 m`
`Delta L = (TL)/(AY) rArr T = (AY Delta L)/(L)`
Fundamental frequency of string
`f = (1)/(2L) sqrt((T)/(mu)) = (1)/(2L) sqrt((T)/(sigma A))`
`= (1)/(2L) sqrt((AY Delta L)/(L sigma A)) = (1)/(2L) sqrt((Y Delta L)/(sigma L))`
`(1)/(2 xx 1) sqrt((Y)/(10^(4)) xx (0.02)/(1)) = (1)/(2) sqrt(2 xx 10^(-6) Y)`
`= sqrt((10^(-6) Y)/(2))`
Fundamental frequency of open pipe
`f' = (v)/(2l) = (292)/(2 xx 0.5) = 292 Hz`
By decreasing stress `(T//A)` in wire, beat frequency decreases i.e. `f gt f'`
`f - f' = 8 sqrt((10^(-6) Y)/(2)) - 292 = 8`
`sqrt((10^(-6) Y)/(2)) = 300 rArr Y = (2 xx (300)^(2))/(10^(-6)) = 18 xx 10^(10) N//m^(2)`