Find the ratio of the fundamental tone frequencies of two identical strings after one of them was stretched by `eta_(1)=2.0 %` and the other, by `eta_(2)=4.0 %`. The tension is assumed to be proportional to the elongation.

Explanation

In case of a stretched string
`n = (l)/(2l) sqrt(((T)/(mu)))`
Where p is the mass per unit length.
IfMbe the total mass ofthe wire, then
`n = (1)/(2l) sqrt(((Tl)/(mu)))`
It should be remembered that when the wire is stretched, the total mass of wire M remains constant.
If original legnth was l, then after elongation
`l_(1) = 1.02 l` and `l_(2) = 1.04 l`
Given that tension `infty` elongation
Hence `(T_(1))/(T_(2)) = (eta_(1))/(eta_(2)) = (2)/(4) = 0.5`
The new frequencies are
`n_(1) = (1)/(2(1.02l) ) sqrt[[(T_(1)(1.02l))/M]]`
`n_(1) = (1)/(2(1.02l) ) sqrt[[(T_(1)(1.04l))/M]]`
Dividing equation-(6.194) byequation-(6.193), we get
`(n_(2))/(n_(1)) = sqrt[[((1.02)/(1.04))((T_(2))/(T_(1)))]] = sqrt[[(1.02)/(1.04) xx (1)/(0.5)]]`
= 1.4