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`.

Answer

`Y = 1.76 xx 10^(11) N//m^(2)`

Explanation

The fundamental frequency of open pipe is `n = (v)/(2l) = (292)/(2 xx .5) = 292 Hz`
Let `L` be the initial length and `L + DeltaL`
the stressed length and `L + Delta = .98 + .02 = 1m`
The fundamental frequency of stressed wire `n = (1)/(2(L + DeltaL)) sqrt((T)/(pir^(2)d))`
where `r` is the radius and `d` is the density of wire.
The stressed wire and the organ pipe (frequency `292 Hz`) produces `8` beats
`:. n = 292 +- 8`
By decreasesing tension velocity decreased At the same time number of beats also decreases.
`:. n = 292 + 8 = 300`
by eqn. we have `(T)/(pir^(2)) = 4 n^(2)d (L + DeltaL)^(2)`
`Y=((T)/(pir^(2)))/((DeltaL)/(L)) = (4n^(2)d(L+DeltaL)^(2))/(((DeltaL)/(L))) = (4xx(300)^(2)xx10^(4)xx1^(2))/(((.02)/(.98))) = 17.64 xx 10^(10) N//m^(2)`